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APPLED MECHANICS 



WOODWARD 



FIRST YEAR WORK 



IN 



APPLIED MECHANICS 



BY 

CALVIN M. WOODWARD 

Professor Emeritus of Mathematics and Applied Mechanics 
Washington University, St. Louis. 



[SECOND YEAR WORK IS NOW IN PRESS.] 



OCT. 1, 1912, 



NIXON-JONES PRINTING CO. 

ST. LOUIS, MO. 



, 






A 



Copyright, 1912, by 

Calvin M. Woodward 



ALL RIGHTS RESERVED 



Hie 

CCI.A320724 



PREFACE. 



This book is written primarily for students entering upon their 
second collegiate year. Their knowledge of mechanics is limited to 
what was gained during their study of physics and from practice in 
laboratory and shop ; hence no apology is needed for making matters 
plain and easy in the first chapters of a work which later on leads up 
to the Theory of Structures, Internal Stress, Motion under Complicated 
Conditions, Forms of Energy, and higher Graphics. 

It is assumed that at every stage of progress the student can by 
reasonable effort read the book understandingly by himself (always 
with pencil and paper at hand), so as to be ready to grasp the condi- 
tions of concrete or ideal problems brought up in class, and follow state- 
ments made in the language of mathematics and mechanics. In other 
words, he must here learn to translate the language of the world of mat- 
ter, motion, and force into mathematical forms. 

The book is not flooded with easy examples which involve nothing 
new. The use of Graphical solutions is a characteristic feature. The 
infinitesimal Calculus is not introduced until Centroids and Centers of 
Action are reached in Chapter VII. In Chapter IX, fiber stress in a 
Bent Beam is introduced in order to fully explain the origin and use 
of the Moment of Inertia of surfaces. The comprehensive study of 
beams will follow in Chapters XIX-XXI. 

The comparatively easy matter of direct Stresses in Framed Struc- 
tures belongs in one's first year in Mechanics. Static Triangles and 
Static Polygons are preferred to abstract equations, tho the latter are not 
excluded. 

The titles of subsequent chapters sufficiently indicate their content. 

The entire volume will be ready for delivery before Christmas, 
1912. 

Notice of errors, whether by the author or the printer, will be thank- 
fully received. 

""•aSLSrSfeiBi. Calvin M ' Woodward. 



TABLE OF CONTENTS- 

CHAPTER. PAGE. 

I. Introduction 1 

II. Co-Linear Forces. The Theory of Couples .... 10 

III. Parallel Forces in a Plane. The Theory of Levers. 

Parallel Forces in Space 22 

IV. Co-Planar Converging Forces. The Static Triangle and 

the Static Polygon. Graphical Solutions of Problems 45 
V. Non-Concurrent Co-Planer Forces. Four Methods of 
Finding the Resultant of a System of Forces: (a) 
Graphic, Analytic, Semi-Graphic, by the Chain or 

Arch Potygon 66 

VI. Converging Forces in Space. The General Case of 
Forces Acting at Several Points in Space and in 
Different Directions 73 



Si 



chapter. Table of Contents Continued. page, 

VII. Centroids of Plane Surfaces, and Centers of Action of 

Uniformly Varying Forces. Hydrostatic Pressures 80 

VIII. Centers of Gravity of Solids 94 

IX. Moments of Inertia of Plane Surfaces. Sections of Built- 
up Beams and Columns 105 

X. The Moments of Inertia Continued. The Relation of 
Strength to M. I. Maximum and Minimum Values 

of I. Comparison of Designs 124 

XI. Direct Stresses in the Members of Frames. Graphical 

Statics Applied to Roof and Bridge Trusses . . . 136 

SUBSEQUENT CHAPTERS IN PART II. 

XII. The Composition and Resolution of Internal Stresses. 
The Ellipse of Stress, and its Use in the Solution of 
Special Problems 

XIII. Kinetics. Translation of Solid Bodies under the Action 

of Constant, Unbalanced Forces. General Formulas 

XIV. Translation under the Action of Varying Unbalanced 

Forces. The Simple Pendulum 

XV. Moments of Inertia of Solids. Rotation under the Ac- 
tion of Constant, Unbalanced Moments. Translation 
and Rotation Combined. General Formulas. The 

Compound Pendulum 

XVI. Deviating Forces. Normal Accelerations. Projectiles 
i XVII. Kinematics. Rolling Curves, Instantaneous Axes. Sim- 
ultaneous Rotations combined. Helical Motion . . 
XVIII. Work and Energy. The [ Conservation of Energy, 

Kinetic and Potential 

XIX. Flexure. The Deflections of Beams variously Supported 

and Loaded 

XX. Graphical Representations of Shear and Bending 
Moments in Beams. Graphical Determinations of 
Moments, Slopes, and Deflections by means of Areas. 

Influence Lines 

XXI. The Distribution of Shearing Stress in Beams and 

Shafts. The Angle of Torsion. Harmonic Oscillation 

XXII. Beams of Uniform Strength. Re-enforced Concrete. 

Continuous Girders 

XXIII. The Deformation of Beams, Struts and Ties. The Doc- 

trine of Least Work. The Solution of so-called Inde- 
terminate Problems in the case of Redundant Members 

XXIV. Miscellaneous Problems. The Stability of Founda- 

tions, Chimneys, and Retaining Walls. The Strength 

of Hanging Cables and Chains 

XXV. The Strength of Columns, Short and Long, under the 
Action of Eccentric Loads, or Deflecting Forces . . 
XXVI. The Energy of Streams of Water and Air. Impact . 
XXVII. The Behavior of Perfect Gas under changes of tempera- 
ture and Pressure. The Efficiency of Compressed Air 



APPLIED MECHANICS. 

CHAPTER I. 

Introductory. 

1. Force is an action between two bodies. It is not a tendency, 
it is a real thing as experience readily shows. A pushes or pulls B, 
and at the same time B pushes or pulls A', there are two bodies and one 
action or force. There can be no force unless there be at least two 
bodies. The bodies may be at rest or they may move. There may 
be an action, a push or pressure, between two bricks in a wall which 
does not move. There may be a pull or tension between two cars 
which are in motion. 

1. In Applied Mechanics the general surface of the earth is assumed 
to be at rest. In Celestial Mechanics full account is taken of the 
motions of the earth and of other heavenly bodies. 

All forces are more or less distributed, either over surfaces or thru 
volumes. The pressure between one's foot and the floor, or between 
steam and a piston, is distributed over the surface of contact; in one 
case unevenly, in the other uniformly; in one case without motion, 
in the other with motion. The earth, in a most mysterious way, 
pulls upon every particle of matter: — an apple hanging on a tree, a 
brick in a wall, and an iron ball flying thru the air; and of necessity 
the apple, the brick, and the ball pull the earth. This is called the 
"force of gravitation." Of all the forces with which mechanics 
deals, gravitation is the most common, and the most inexplicable. 
We know, however, that nothing can escape it, and that it is not 
sensibly affected by such distances as are generally considered in this 
book, or by intervening objects. 

2. It is highly important that the student regards the force of 
gravitation as a pull, distributed thruout the whole volume of a body. 
Moreover the action (or pull) of gravitation upon a body must not 
be confused with the action between the body and the platform 
or foundation upon which the body may rest. When a body like a 
block of stone rests upon the ground, there are two actions between 
the stone and the earth, viz: a pull down and a push up, which 
exactly balance, one action being distributed thru volumes, the other 
being distributed over the surface of contact. 

3. The plural word "volumes" is used advisedly. One volume 
is that of the stone which is made up of an infinite number of particles; 



INTRODUCTORY 



the other volume is that of the earth which also consists of an infinite 
number of particles; and there is an action between every particle 
in the stone and every particle in the earth. What we call the weight 
of a body is the resultant of all the separate pulls, and its direction 
is towards and away from the earth's center. The stone pulls the 
earth just as much as the earth pulls the stone. See "Attraction" 
in a later Chapter. 

2. When Sir Isaac Newton, the master mind in Mechanics, stated 
as a fundamental law that action and reaction were equal and opposite, 
he meant only this: That when there is an action between A and B, 
the measure of A's action on B is exactly equal to the measure of B's 
action on A, and in the opposite direction. This sounds as though 
there were two forces for one action, but such is not the case. There 
is only one force, but its action may be viewed from the standpoint 
of A y or from the standpoint of B. That is, if we are considering the 
condition or behavior of A as regards rest or motion, we only take 
into account the action of B upon it, and we give that action its 
proper direction; if, however, we are considering the status of B, 
we take account only of A's action upon it. 

1. This law of Newton does not explain the stability of the block 
of stone which rests on the ground in a former illustration. In that 
case there were two actions between the earth and the stone, a pull 
and a push, that is, an attraction and a pressure; and the pull and 
push balanced. 

It is true in the case considered, that if there were no pull, there 
would be no push, because we have assumed that the stone was at 
rest, and hence the two actions must balance. If there was but one 
action, the stone would not be at rest; this must be made clear. 

2. Suppose a stone to be suspended 
by a derrick, and to be at rest. There 
is now an action between the stone 
and the earth; the earth is pulling 
down on the stone and the stone is 
pulling up on the earth; that is one 
case of action. In the next place there 
is an action between the stone and 
the derrick; the stone is pulling down 
on the derrick, and the derrick is pulling up on the stone — this is a 
second case of action. Thus we see that the stone is acted upon 
by two bodies: the earth which pulls down and the derrick which 
pulls up, and as the stone is at rest we know that these two pulls are 
equal in magnitude and exactly opposite. 




Fig. 1 



MECHANICAL UNITS 3 

Now suppose the rope or chain thru which the derrick acts, is cut 
or breaks. The action between the stone and the derrick ceases, and only 
the action between the stone and the earth remains. Hence it cannot 
be at rest; and the stone and the earth begin at once to approach each 
other. However, the earth's motion is so immeasurably small that we 
leave it wholly out of account, and merely say: "the stone falls." 

3. Stress. Thus far we have thought of A and B as two separate 
bodies, but it is evident that they may be separate parts of the same 
body like adjacent strata in the earth, adjacent leaves in a book, 
adjacent links in a chain, adjacent particles in a steel rod or a con- 
crete post. When A and B are adjacent parts of a continuous body, 
their mutual action, no matter what its character may be, is called 
Stress. If it be a pull it is called tensile stress, and the continuous 
body, rope, wire, rod or bar is said to be in tension. If the adjacent 
layers press against or upon each other in a post, strut, block or wall, 
the post, strut, block or wall is in compression, and the stress is 
called compressive stress. 

The link of a chain illustrates fairly well adjacent parts of a con- 
tinuous body. Suppose in Fig. (1) that the stone is suspended to 
the derrick boom by means of a chain. The chain is in tension and 
every link is acted upon by two forces (independently of the earth's 
pull or attraction upon the material in the link) the downward pull 
of the link below it, and the upward pull of the link above it. Further- 
more, it is evident that the upward pull of the topmost link must 
equal in magnitude the downward pull of the lowest link, plus the 
pull, or attraction, of the earth upon all the material in the inter- 
mediate links. 

4. Magnitude of forces. Units. We are immediately conscious 
that forces or actions have magnitude; one pull or attraction is greater 
or less than another; one push, pressure or repulsion is greater or less 
than another. In order to express magnitude with precision we must 
have a well known unit of force with which all other forces may be 
compared numerically. A certain pull or push shall be called one, 
and like other units in common use, it shall have a name. The unit 
of time in mechanics the world over is a second (or a multiple of a 
second); the unit of length or distance most commonly used in the 
Anglo-Saxon countries is a foot (or a multiple thereof), though the 
meter, a French unit, is in common use in physical text-books and 
laboratories. The unit of value in North America is a dollar. In 
all these cases we know by observation, experience and frequent 
use just what these units are, and what a given number of seconds, 
feet, meters or dollars means. 



4 INTRODUCTORY 

5. Units of force. When, however, we come to the unit of force 
and its name, we find great diversity in the text books, lecture rooms, 
and laboratories of every engineering school in the land, in spite of 
all efforts of the users to bring about the adoption and general use of 
a common unit. 

1. The first and most common is the avoirdupois pound by which 
we measure the weights of all sorts of things, building materials, 
foods, crops, ores, manufactured articles, etc. (i. e. y the pull of the 
earth upon them), the tensions in ropes, cords, chains, drawbars, 
rods and eye-bars; and the pressures of gases (air, steam, etc.), and 
of liquids (water, oils, etc.). 

2. The second unit of force is the poundal. This unit of force 
is about one-thirty-second of an avoirdupois pound, or half an ounce. 
Its use is extremely limited, but is not unfrequently found in works 
on applied electricity, where it measures the electrical forces of attrac- 
tion and repulsion. 

3. The third unit of force is a kilogram, or its one-thousandth part 
called the dyne. This unit of force is of French origin, but its use, 
like that of the meter, is world-wide in physics, chemistry and engi- 
neering. In magnitude the Kilogram is about 2.2 pounds. 

G. The meter (which was intended to be one ten-millionth of a 
quadrant of the earth's meridian, and which is approximately so), 
is really the distance between two engraved marks on a platinum- 
iridium bar, carefully preserved at Paris. An accurate copy of this 
bar is in the archives at Washington. The meter bar was received 
by the President of the United States in 1890; the entire metric system 
is now legalized in all the territory under our flag. The foot, con- 
sisting of 12 inches, is defined as a certain part of a meter, shown by 

the relation that _ . ark a „ . . 

1 meter = 39.37 inches. 

1. At the same time in 1890 a standard block of metal (platinum- 
iridium) was received and deposited at Washington, which, at the 
level of the sea, and at 45° north latitude was declared to weigh one 
kilogram (kg.). Another block, under the same conditions, is declared 
to weigh one pound (lb.). The relation is very exactly 

One kilogram = 2.20462 lbs. 

2. The relation of the kilogram to the meter is shown by this: 
A kilogram is the weight, at the level of the sea, latitude 45°, of one 
one-thousandth part of a cubic meter of distilled water at a tempera- 
ture of 0° centegrade, or 32° Fahrenheit. Accordingly, a cubic meter 
of water weighs 1,000 kgs. under standard conditions. 



IDEAL CONDITIONS AND IDEAL PROBLEMS 5 

3. It will be well for the student to fix these standard relations 
well in his mind as he must think readily in terms of each. 

1 meter = 39.37 inches. 

1 foot = 0.3048 meter. 

1,000 meters = 1 kilo-meter = § mile, nearly = 3281 feet. 

1 kilogram = about 2.2 lbs. 

1 ton = 2,000 lbs. = 907.2 kgs. 
For greater precision (rarely necessary) see an encyclopedia. 

•7. Units of mass and time. There are in every system of units 
a unit of time and a unit of mass, which will be fully explained when 
we come to expound the laws of motion and moving bodies. 

Meanwhile, the student must not be confused or disturbed by the 
use in text books and scientific journals of such expressions as "a 
mass of so many pounds," or "a mass of so many kilograms"; the 
meaning merely is: — a quantity of material upon which the earth's 
attraction is so many pounds weight, or so many kilograms. The 
simplest way to find the magnitude of the earth's pull or attraction 
upon a given quantity or mass of material is to weigh it by a spring 
balance, a process which need not be explained. The essential thing 
is to bear constantly in mind that the earth's pull upon a mass is 
always & force, and that that force should never be confused with the 
material pulled or acted upon. 

8. Mechanical problems. 1. Problems in mechanics are always 
given under conditions which are more or less ideal, i. e., not real. 
For instance, in the case of the stone hanging upon the boom of a 
derrick, it was assumed that the action of the air upon the stone 
was either nothing at all, or that its action was self-balanced, and 
yet we know that such could not be the fact. We know that the 
pressure of the air upon every square foot of the surface of the stone 
is about a ton. We also know that the upward pressure of the air 
upon the bottom or lower surface is a little greater than the down- 
ward pressure upon the top or upper surface; consequently the total 
up-and-down action of the air is not self-balanced. The difference 
or buoyancy is, however, so small compared with the other forces 
acting (the pull of the earth and the pull of the rope or chain), that 
it was neglected, and the magnitude of the earth's pull was assumed 
to be equal to the pull of the rope. 

£. Again, there are always air currents, which are ignored, unless 
those currents are very strong, as in the case of high winds. In like 
manner we shall often assume that solids and gases are of uniform 
density tho we know that they always vary more or less. We 
shall at times suppose that the surfaces of solid bodies are smooth, 



6 



INTRODUCTORY 



tho we know none that are perfectly smooth; and that the earth's 
pull on a body when a few feet above the sea level, or a few feet below 
it, is just the same as it is at sea level; and that change of latitude has 
no effect upon the earth's attraction, tho it is well known that strictly 
speaking such is not the case. Sometimes we shall assume that bodies 
are perfectly rigid tho we are positive that no bodies are strictly rigid. 
3. Such false assumptions do not render the problems and their 
solutions useless. The solutions are just as accurate as are the as- 
sumptions. If the latter were sufficiently accurate, or "near enough," 
then the solution is near enough. Both the student and the teacher, 
however, are warned against the danger of ignoring conditions which 
are so far from real conditions, that the results of solutions are very 
misleading and mischievous. It is rarely wise to ignore friction, 
internal stresses and elasticity. Even with the best assumed con- 
ditions, problems are still somewhat ideal. In the problems which 
are solved or given for solution in this book, care will be taken to have 
the assumed conditions approximate closely real conditions, so that 
the conclusions reached may have value, and serve as guides for 
future conduct and use. 

9. The recognition of forces. Fig. 2 should represent a heavy 
block resting on a table. This block supports a smaller block and a 

man who holds a loaded basket on his 
head. Partial support is given to the 
large block by a "flexible and weight- 
less" rope passing round the large block 
at the lower end and after going over a 
"smooth" peg, sustains a weight of 40 
lbs. at the upper end. The student is 
to give the number, character, ''sense" and 
magnitude of the forces or actions upon 
each individual body: A, B, C, Z), E and 
P. Character shows whether it is a push 
or a pull, and how it is distributed. 
"Sense" tells the direction of the action 
of a force on the body just then under 
consideration. 

For example, the Man is acted upon 
by three forces: 

1. The downward pull of the earth's 
attraction of 150 lbs., which is distributed 
thru the volume of his entire person. 

2. A downward pressure of 20 lbs. from the loaded basket distributed 
over the upper surface of his head. 




Pig. 2 



THE RANGE OF COMMON SENSE 7 

3. An upward pressure of 170 lbs. from the large block distributed 
under the soles of his shoes. 

The student should write out similar statements for the other 
bodies in the problem, paying special attention to his language as well 
as to his thought. 

10. Ideal conditions. Several matters will be noticed which re- 
quire mention. We have assumed several conditions which are more 
or less erroneous. 

1. We have left the air entirely out of consideration whatever its 
action may be. 

2. We have assumed that the peg over which the rope passes is 
perfectly smooth. 

3. That the rope itself, tho strong, is perfectly flexible and im- 
ponderable, so that the tension in it is the same (viz.: 40 lbs.) from 
end to end. 

In so-called practical problems, it is generally said that such trifles 
are unimportant, but it is easy to see that they might not be trifling 
or unimportant. 

4. Finally, we must infer some things which the statement of the 
problem does not give. This inference is the result of a process of 
reasoning of which one is hardly conscious. How did we know that 
the large block pressed or lifted up against the man's shoes with a 
force of 170 lbs.? Had both man and block been either descending or 
ascending faster and faster, the action between shoes and block would 
not have been 170 lbs., but since both were standing still, we know 
that the forces acting on the man must balance. Hence the only 
upward force must alone be equal to the sum of the two downward 
forces. 

5. Such reasoning we call "Common Sense," but it is strictly in 
accordance with the axioms or laws of mechanics. It will be seen 
later on that the range of common sense can be greatly increased as 
the laws and conditions which obtain in the interaction of bodies, at 
rest and in motion, are known and understood. 

11. The transmission of forces. In the case of a chain under 
tension, it is easy to see how a pull upon the link at one end is trans- 
mitted, or passed along, _^ ^^ 

thru all the links to the < ^J^>>Hooo«»aocc^^ > 

other end. Fig. 3 

Accordingly, we say that 
there is an action between A and B by means of the chain C. Still 
more simply, we may say, when we are considering the status of B y 
"A is pulling B." It is evident that the length of the chain is not 



/ 2 



INTRODUCTORY 



¥ 



1 ljl Ijl Ijljljl ^J^_fl 



of any account so long as the chain, like the rope in the problem, is 
without weight, and therefore straight. 

In like manner a body A may act upon or push a body B, by 
means of a strut, post, brace or leg. We shall see later that a force can 
be transferred or transmitted sideways (laterally) thru solid bodies. 

12. Forces acting at point. In the statement and solutions of 
ideal problems in mechanics, perhaps the most striking assumption 
is, that finite forces act at points. Now all forces in reality act at 
surfaces or thru volumes, yet a point has neither volume nor surface. 
Nevertheless the notion of concentrating a distributed action or force 
is most natural and very convenient. Common experience, and 
therefore common sense, convinces us that a trap door, a platform, or 
a warehouse floor, if sufficiently rigid, could be supported by a single 
prop or post, provided it be put in the proper place. 

I Imagine a stiff plank, loaded 

with bricks, balanced upon the 
top of a small square post. This 
is easily done in fact, and still more 
easily imagined. Now the earth's 
attraction upon plank and bricks 
must be exactly balanced by the 
lift of the post; hence the total 
downward action must somehow 
be concentrated, or transmitted, 
to a point or small area just above top of the post. Similarly, the 
upward action of the post may be centered within a small area im- 
mediately in contact with the surface of action of the plank. These 
two areas may ideally be made as small as we please, i. e.> each may 
be called a "physical point" and the two total actions may be 
thought of as acting at a common point. 

13. The graphical representation of a force. It will now be easy 
to represent an action upon a body (from any source whatever) by a 
straight line or arrow, the direction in which the arrow points show- 
ing the "sense" or direction of the action; the length of the line drawn 
to scale (so many pounds or tons to the foot, or so many kilograms to 
the meter), showing the magnitude of the force or action; and the 
position of the line showing the "line of action," any point of which, 
in or on the body acted upon, may be taken as the physical "point 
of action" of the acting body. In figure 4, the upper arrow shows 
the direction, magnitude and position of the earth's total attraction 
on the plank and bricks; and the lower arrow shows the direction, 
magnitude and position of the upward action of the post upon the plank. 




Fig. 4 






ABSTRACT PROBLEMS 9 

14. Abstract forces. Altho no force can exist, except as an 
action between bodies, it will be convenient, when we study the con- 
centration (composition or combination) of separate or distributed 
forces, to leave the description of the bodies whose actions we are 
considering, wholly out of our thought, and treat the forces as inde- 
pendent things; that is, we withdraw (or abstract) the forces from the 
bodies creating them, and reason about them abstractly, just as one 
might add six dollars to eight dollars, thereby producing fourteen dollars 
without having any money at all. So we may say: "Suppose we have 
a force fully represented by the arrow line AB, and, in the same plane 
with AB, another force fully represented by the arrow line CD. Our 
problem is now to find a single force XY whose action shall in all 
essentials represent, or be equal to, the combined action of the two 
given forces." 

In the solution of this problem, we take no thought of the bodies 
producing the given actions. We care not what they are, just where 
they are, nor why they act, except that we must know their relative 
positions. Of course, we assume that they act upon the same rigid 
body, but we do not care in this problem what that body is. In other 
words our problem is purely abstract. 

1. Thruout this book we shall mingle abstract problems freely 
with those which approximate more or less closely real ones. With 
abstract problems, we shall aim to illustrate general methods, estab- 
lish general laws, and derive general formulas, and then apply them 
to the solutions of problems derived from all sorts of sources and con- 
ditions, of a practical character. The number of possible problems 
is infinite; many are fanciful and useless; many are too difficult for 
our range of mathematics; many are intensely practical and admit of 
easy solutions. In our analysis we shall draw freely upon the student's 
knowledge of Algebra, Geometry, plane and solid, and Trigonometry; 
and later on we shall assume Analytic Geometry and Calculus. The 
more thoroughly these have been mastered the better, but the more 
difficult operations will be clearly explained. 



PART I. 

STATICS. 
CHAPTER II. 

The Combination and Resolution of Forces. 

STATICS is that department of Mechanics in which all the bodies, 
acting and acted upon, are at rest or stationary. A body at rest 
under the action of two or more other bodies is said to be in Equilib- 
rium; and the forces acting upon it are said to Balance. 

15. Forces having a common line of action. The simplest 
possible example is that of two forces which exactly balance or neutral- 
ize each other. It is evident without any attempt at proof that they 
must not only act along the same line, but they must have equal 
magnitudes and opposite directions. Under no other conditions can 
two forces balance. 

1. In so far as the state of rest or motion of a rigid body is con- 
cerned, two balancing forces may be left out of account, while we are 
considering the actions of other forces upon the same body. This 
cancellation of two balancing forces, is analagous to the striking off 
of equal terms in the members of an algebraic equation. 

2. Conversely, we may at any time, to facilitate a solution, 
introduce two balancing forces, representing the actions of two imagin- 
ary bodies upon the body we are considering, without in any way 
affecting its state of rest or motion. 

3. When several forces, of different magnitudes, have a common 
line of action, while some act in one direction and some in the other, 
they are readily reduced to two by adding those which have a common 
direction, and then reducing the two to one by allowing the smaller 
to cancel or neutralize an equally large part of the other; or the cancel- 
lation and addition may be carried on simultaneously. All this is 
readily expressed in the language of algebra. 

Let F be the numerical measure of the magnitude of a force, and 
let one direction along the common line of action be considered as 
positive, and the other direction negative. 

T>F represents a process of algebraic addition of as many numbers 
(of units of force) as there were of assumed forces, acting upon a 
(real or imaginary) body. Finally, if we represent the magnitude 
(10) 



CO-LINEAR FORCES 11 

of the single force which results from the addition by the letter R, we 
have our first equation in mechanics. 

ZF = R (1) 

If the forces under consideration actually balance, the value of R 
is zero, and the equation becomes 

2F-0 (2) 

16. Conversely. Since forces having a common line of action may 
be combined to form a single force, so a force may be resolved or 
separated into any number of component forces having the same line 
of action and preserving the algebraic sum. 



F5 F3 Fl ^ F , 2 F A F6 F7 F8 



Fig. 5 

The student should not fail to see the full meaning of these 
equations. An illustration will surely be of value. Fig. 5 is to rep- 
resent a "tug of war" between a team of three men and a team of 
five boys. The three men have hold of one rope and are pulling to 
the left. The boys have another rope and they pull to the right. 
Both ropes are attached to a pin carrying a flag. Ropes, pin and 
flag are "imponderable," i. e. 9 their weight is not taken into account. 
Of course the effort of each team will be to pull the pin and flag its 
way. As soon as the signal is given to "Pull!" eight bodies (boys and 
men) act (thru the mediation of the ropes) upon the body (pin) A. 
Let us assume that forces acting towards the right are positive and 
those to the left are negative, and that 

F 8 =+200R> F 2 =+180ib F 4 =+175lb ^=-306 lb 
2^3= -284 lb F 6 =+140lb F 2 =+150lb F b = -255 lb 

Now in this case 2F means 

+ 180 + 140 + 200-284 + 175 + 150-306-255 

and since the algebraic sum is zero, 

2F = 0. 

Hence the forces balance, and that the pin A stands still. 

If on a second pull 2F<0, the men win; if 2F>0, the boys win. 

1 7. An action in Statics does not produce motion ; on the contrary 
an action prevents motion. A brick in the wall is acted upon by the 



12 



THE THEORY OF COUPLES 




Fig. 6 



brick above it, and the brick below it, and by the earth which attracts 
it, but it does not move, because the forces balance each other, that 
is: the action of the lowest brick is equal and opposite to the com- 
bined action of "gravitation" and the uppermost brick. 

1. When forces do not balance, and motion is caused or modified, 
we have a problem in Kinetics, which will be discussed in later chapters. 

18. Couples. When a body is acted upon by two other bodies 
in such a manner that the two forces have equal magnitudes, different 
but parallel lines of action, and opposite directions, they cannot 
balance, neither can their joint actions have a tendency to move the 

body up or down, to the right or to the 
left. Their sole tendency is to make the 
body turn, or, if it be already turning, to 
modify that turning. 

1. Suppose a piece of cork to be float- 
ing upon a tub of water. The pull of the 
earth upon it is exactly balanced by the 
upward pressure of the water under it so 
that it is at rest, and we may ignore these 
two balanced forces. Now suppose that 
two needles are prest into the upper sur- 
face of the cork at a and 6, Fig. 6, and that two parallel threads 
are stretched to the edge of the tub in a plane parallel to the surface 
of the water, as shown. Suppose that at A and B gentle but equal 
pulls upon the threads are given simultaneously. Instantly the cork 
begins to turn "right handed." These two new forces form a "right- 
hand couple," whose tendency is obvious. 
The perpendicular distance between the 
two lines of action is called the arm of the 
couple. 

2. Next suppose the cork at rest again 
and that two more needles are put in at c 
and d, Fig. 7, and that threads are carried 
across and fastened at C and D, so that 
they are parallel to each other and in a 
plane parallel to the surface of the water. 

If now the equal pulls at A and B are 
again applied, the cork stands still, and it will be found that the 
threads cC and dD are in tension. Moreover, since the cork does not 
move either towards C or towards Z), the tension in cC and in dD are 
equal; they thus form a "left-hand couple." We consequently have 
two couples which balance each other. It is evident that the turning 




Fig. 7 



EQUIVALENT OF COUPLES 



13 



effect or tendency of a couple must depend partly upon the common 

magnitude of the two forces, and partly upon the distance between the 

two lines of action. In short, the magnitude of a couple's influence is 

measured by the product of one of its forces by its arm. This product 

is called a "moment," and it will be represented by the capital letter 

M, so that ,, ^ 7 

M = Fl 

This will be proved when we come to resolve a couple into Fl unit 
couples; SO. 

3. The student must bear in mind that F in this equation is the 
number of units of force in one of the given forces, and I is the number 
of units of length in the given distance (between the parallel lines of 
action), and that therefore M is the number of units of moment. 

When we speak of "the force of a couple," we shall mean the common 
magnitude F. 

19. The unit moment is not a fundamental unit, but it is derived 
from two which are fundamental: the foot and the pound; hence a 
moment is read as so many "foot-pounds, foot-tons," etc. 

The word "foot-pound'' 'as here used must not be mistaken for a 
later use of the same word as the unit of work. The moment unit 
expresses and measures the turning action of a body, but its action 
does not require actual motion. An unbalanced moment would 
turn a body; but in all problems in Statics, moments are balanced 
and no motion results. 

20. The Freedom of a Balancing Couple. Referring again to Fig. 
7, it is evident that the needles c and d could have been placed any- 
where on the cork, and that the threads drawn straight from them 
could have been made fast anywhere on the circumference of the tub, 
provided only that the threads were separate and parallel, and that 
they were in a horizontal plane, so as to be 
prepared to act as a left-hand couple. It is 
important that the student see that these 
conditions are sufficient as well as necessary. 

1. Suppose the body B, Fig. 8, to be acted 
upon by the two forces Fi and F 2 forming a 
right-hand couple with the arm I. It is evi- 
dent that the "point of application" of F can 
be anywhere on the line of action, in each 
case. 

£. The direction of the two lines of force can be any direction in 
the horizontal plane of the couple. For suppose new points of attach- 
ment are taken and a new direction to the parallel lines be chosen 




Fig. 8 




14 THE THEORY OF COUPLES 

as shown by F 3 and F 4 , so that both force and arm remain 
numerically the same (leaving the couple still right-handed) the 
moment must still be M = Fl. 

3. The turning effect with reference to any 
point in the plane of the couple taken as an 
axis is always the same. Let a vertical at 
Fig. 9 be taken as an axis about which the 
couple M (i. c, the two forces with the arm 
/) has a tendency to turn the body B. The 
tendency of Fi taken alone is evidently meas- 
ured by + F(c-\-l), and the tendency of F 2 
is obviously — Fc. The tendency of the couple 
is then the sum of the tendencies of its parts, viz.: 

F(c + l)-Fc = Fl 

The arbitrary distance c has disappeared from the equation and 
there is nothing in the result to indicate where was taken. This 
means that whether c be long or short, positive or negative, the 
result is the same. 

2 1 • Hence we conclude, that a couple acting upon a rigid body can 
be shifted at ivill in its own plane to new points and directions without 
affecting its turning tendency provided only that its moment and sign 
be unchanged. 

Since a given couple can be balanced by one of any number of 
left-hand couples constituted as above described, these left-hand 
couples must have the same moment and be equivalent to each other. 

22. We therefore conclude in general that two couples are equiv- 
alent which have the same moment, which lie in the same plane and have 
the same sign (that is they tend to turn a body in the same direction). 
It follows that a couple F\h may be replaced by a couple F 2 l 2 provided 
F 1 l 1 = F 2 l 2 , as: (10 lb) X (6 feet) = (4 lb) X (15 feet); = (20 lb) X (3 feet). 

23. The moment of a force. If the axis be taken in the line of 
action of one of the forces of a couple, the moment of the couple is 
identical with the product of the other force by its distance from the 
axis. 

Definition. The product of a force, by its distance from a given 
point (or axis) is called the Moment of that force with reference to that 
point or axis. 

24. The turning effect or tendency of a couple is not changed by 
a shifting of it to a parallel plane. 

Proof. Suppose we have two couples M i and M 2 equal in all respects, 
but acting in parallel planes H and K. Suppose in the plane H, a 



THE AXIS OF A COUPLE 



15 




Fig. 10 



third couple M 3 equal to M\ in all respects except the direction of its 
turning tendency; it may be so shifted as to completely neutralize 
Mi and hence balance it. It will now be shown that it can equally 
well balance M 2 in the parallel plane K. Let M 3 be so shifted that the 
arms of M 3 and M 2 are parallel, when M 3 becomes M 3 as shown in 
Fig. 10. It is now easily 
seen that the four equal 
forces F 2 , F 3 , F 2 ', F 3 ' 
balance, for, by sym- 
metry, the concentra- 
tion of F 2 and F 3 is a 
parallel force of 2 F 
thru 0, which is mid- 
way between them. In 
like manner the concen- 
tration of F 2 and F 3 
gives a parallel force of 
2 F' thru 0, in the 

direction opposite to 2 F, so that the resultant is zero. Hence M 2 
and M 3 balance. Now since Mi and M 2 can separately be balanced by 
the same couple, they must be equivalent. 

25. Definitions of the "axis of a couple" and of positive and 
negative couples. 

It has been seen that the point (see Fig. 9), assumed for the pur- 
pose of computing the resulting tendency of a couple, may be any- 
where in the plane of the couple. Now through imagine a line drawn 
perpendicular to the plane of the couple; it will be perpendicular to all 
the parallel planes to which the couple may be shifted. Hence 
may be any point in space, and the perpendicular line will be called 
the "axis of the couple." 

1. This axis is a purely ideal matter; it has no dynamic function. 
In Statics since the forces which act on a body balance, whether they 
form couples or not, the axis of a couple is not a restraining device, 
for it does not react upon the body in any way. It is merely an aid 
in our analysis. 

2. Now as to the sign of a couple. When we stand in front of the 
dial of a tower clock, the hands appear to us to move "right-handed," 
or "clock- wise." To a person in the tower, however, the motion of the 
same hands appears to be left-handed. It is therefore necessary to 
determine in every case what "right-handed" means. To do this, 
take a point as an origin on the assumed axis, and let one end of the 
axis, +X, be called positive, and the other, — X, be negative; and let 



16 



THE THEORY OF COUPLES 



+1 be so taken that the direction from +X to shall be the absolute 
direction in which one must look in order that the couple may appear 
to him to be right-handed or positive. Left-handed couples are 
negative. In the case of the F-axis, look from + F towards 0. 

26. The addition of couples having the same axis. Suppose two 
couples in parallel planes act on the same body. First, let one or 
both be shifted till they act in the same plane. Second, while pre- 
serving the product value of force by arm, let one or both factors be so 
modified that the forces in the two couples shall have a common 
magnitude. Third, let their positions be so changed that a force of 
one couple shall directly balance and neutralize a force of the other 
couple. Fig. 11a shows the two couples in the same plane, and Fig. 116 
shows them so shifted and modified that two forces neutralize each 

other, and only one couple remains 
which has an arm equal to the sum 
of the two arms, and a moment 
equal to the sum of the given 
moments. 

Example — Let Mi have a force of 
4 (units) and an arm of 5 (units); 
and Mi have a force of 6 (units) and 
an arm of 3 (units). 

Let them be shifted and modi- 
then combined into one couple 




(6) 



(a) 



Fig. 11 



fied while still equivalent, and 
Ms = 38. 

1. It is evident that any number of couples with a common axis could 
be added in the same way. A negative or left-handed couple would 
have the effect of shortening the arm of a positive couple. Show this. 

2. An equally simple method of adding co-axial couples would be 
to reduce them all to a common arm, so that when superposed the 
result would be one arm with equal and opposite sets of co-linear 
forces, making a single couple. 



Example. — Given two co-axial couples: 
F = 7 1 ... , F = 12 



F = 7 \ 

1 = 8 j 



positive, and 



1= 2 



negative. 



Reduce to a common arm Z = 4, and combine by superposition. 

26 1 . The resolution of couples. The converse of the last section 
shows at once that a couple may be resolved (or separated) into any 
number of couples by inserting at different points on the arm equal 
and directly opposite forces, each of the given magnitude F, and each 



THE RESOLUTION OF COUPLES 



17 



A 

/ v 


) 


J 


if 




I ^ 


; — dA i 


' % 


^ s 


/» 




y 


v 










(a) 





Fig. 12 



parallel to the given forces. The forces introduced balance each other 
in every pair, and therefore they do not affect the condition of the 
body acted upon. The given couple (F)X(AB), is resolved into 
five component couples, having the same axis, forces of a common 
magnitude, and different arms. 

Still further, every one of the ten 
forces can be resolved into any num- 
ber of components (the same for each 
of the two forces of a couple). 
Thus each of the above five couples 
may be resolved into several couples 
with equal arms, as is the one shown 
in (b). 

It thus appears that the original couple may be resolved into 
any number of component couples. 

When these two methods of resolving a couple into components 
are combined, it is evident that two couples, acting on the same body, 
having the same axis, and equal moments, may be shown to be equiva- 
lent, because they can be resolved into the same number of equal com- 
ponent couples. 

It follows that a couple which has an arm 7 units long and a 
force equal to 10 units has a turning tendency 70 times as great 
as the moment of a couple which has a unit arm and a unit 
force. Hence the turning tendency or moment is measured by the 
product of its force, by its arm, i. e., the product of the two numerical 
measures. This is the proof which was promised in 18. 

The student must never confuse his thought by saying that we can 
really multiply a force by an arm; that would be absurd. What we 
really do is to multiply one number by another number, and the prod- 
uct is still a number. 

What are the "conditions" under which it is proper to say that a 
length of three feet and a force of seven lbs, give us a moment of 21 
foot-lbs.? 

From all this it follows that the essential things about any couple 
are: the direction of its axis, its moment, and its sign. 

27. Couples as vectors. It is now evident that the essential 
things of a couple may be represented by a single straight line or 
vector: It must have length to represent the moment of the couple; 
it must have direction to show the axis, and therefore the plane of the 
couple; it must show the "sense" or tendency by pointing in the 
direction in which the couple must be viewed to appear right-handed. 
Couple vectors will be used in discussions which follow. 



18 



THE THEORY OF COUPLES 



28. The combination of couples which do not have the same 

axis, i. e.y which are not in the same or parallel planes. 

1. Let there be two couples in planes perpendicular to each other, 
and acting on the same body. They can be combined as follows: 
Reduce the couples to equivalent couples so that the four forces will 
have a common magnitude. Next, shift and if necessary turn the 

couples, each in 
its own plane, 
till there are 
two equal and 
directly oppo- 
site forces act- 
ing along the 
line of intersec- 
tion of the two 
planes. These 
two co -linear 
forces balance 
each other, and 
there remain 
two equal forces 
which form a 
couple in a third 
plane which is 
parallel to the 
intersection of 
the first two. 
The arm of this 
resultant couple 

is evidently the hypotenuse of a triangle whose legs are the reduced 
arms of the given couples. 

Illustration, Let OX, OY, OZ be rectangular axes. Fig. 13. Let 
Mi = 24 be a couple in the plane YZ, right handed to one looking from 
-\-X towards 0. Let M 2 = 15 be a couple in the ZX plane, right 
handed to one looking from + Y towards 0. Reduce each couple to 
an equivalent couple with a force 3, and a proper arm; so that 
Jf 1 = 3X8, and M 2 = 3X5. When the reduced couples are shifted to 
the intersection OZ, so that two of the forces neutralize each other, 
there remains only a couple having one force in the plane YZ acting 
down, and the other in the plane ZX acting up, with an arm which 
equals V (64+25); so that the resultant couple is found to be: 

3V (64+25) = V [(24) 2 +(15) 2 ] = M = V (M 1 2 + M 2 2 ) 




Fig. 13 



COUPLES WITH DIFFERENT AXES 



19 



The axis of M is evidently a line in XY (or parallel to that plane) 
perpendicular to the trace T 2 T 2 of the third plane ABCD, on XY. 

Fig. 13 also contains a graphical solution of the problem by the 
method of vectors. 

The direction of M is determined by the equations: 



sin 



6 = 



M 



cos 







The moment Mi is laid off to 
scale on +Z = HO = 24. In like 
manner M 2 is represented fully 
by KO = 15, and then M by the 
diagonal of the rectangle. 

3. The student should prove 
by geometry that the triangles 
RHO and JIO are similar and 
that consequently RO is per- 
pendicular to t 2 t 2 . 




29. When the planes are 
oblique to each other the analysis 
is the same, but the triangle of Fig . i4 
arms is oblique. 1. The algebraic 

solution is as follows: — Let the given couple be reduced to equivalent 
couples, the magnitude of F being common. Then 

M l = Fl u ELndM 2 = Fk 
and M =F (Zi 2 +Z 2 2 — 2ZiZ 2 cos^) 1 

in which <j> is the angle HOK (Fig. 14.) 

2. In like manner the Graphical solution gives M in magnitude 
and shows the direction of the axis which is perpendicular to the plane 
of the new couple. M = V (Mi 2 + M 2 2 -ZMiM 2 cos <£). 

Let both planes be perpendicular to the paper, and let their traces 
intersect at 0. M\ fully represents the couple in OH, and M 2 fully 
represents the couple in OK; and consequently M fully represents 
the resultant couple in the plane TT. 

30. It is readily seen that two couples in non-parallel planes 
cannot possibly balance each other; and it is as easily seen that two 
couples in non-parallel planes may be balanced by a third couple, and 
that the moment of the balancing couple must be equal to M, the 
resultant of the other two, and the arrow line of its axis must be just 
the opposite of that of M .* 

*A resultant couple has been denned as the couple which balances the couple 
which balances the given set of couples. 



20 



THE THEORY OF COUPLES 




31. Three couples balance. It follows from the last two problems 
that when the representative vectors of three couples, are of such 
lengths and directions that they may form a plane triangle, when they 
are drawn in order, as they point, they balance, and the body on which 
they act is in equilibrium. 

Thus the triangle ABC represents fully three couples acting on the 
same body, each with a definite tendency to turn it in a definite direc- 
tion, and yet they balance and (so far as these 
couples are concerned) the body is at rest. The 
triangle ABC is called the moment triangle of 
equilibrium, 

32. The resultant of three or more couples. 

Having found the resultant of two couples as 
shown above, a third couple may be combined with the resultant of 
the two, and a resultant of three be found. This last may then be 
combined with a fourth, and so on for any number of couples. 

1. When the planes of three couples are the rectangular co-ordi- 
nate planes, the problem is simple. The resultant of ifi and M 2 is 
M ' = V (Mi 2 -r-i/ 2 2 ); and the resultant of M r and M z is 

M = V(l/i 2 + M 2 2 + M 3 2 ); 

its vector is the diagonal of a 
rectangular solid constructed on 
Mi, Mi, and M z as confluent 
edges. 

2. If the planes of several 
couples are perpendicular to a 
common plane, the graphical 
solution is most simple. In Fig. 
16 let the plane of the paper be 
the common perpendicular plane. 
Let the planes of the given couples 
have the traces AB, BC, CD, 
BE, EH, etc. From any point (Fig. 16) draw Mi fully represent- 
ing the couple in AB; from the end of the vector Mi, draw M 2 fully 
representing the couple in BC; OQ, then, fully represents the resultant 
of Mi and M 2 . From the extremity of the vector M 2 , draw Mz, 
then if 4 , and so on. OT is the resultant of Mi, M 2 and M z ; and 
OR = M is the resultant of them all. The balancing couple is repre- 
sented by the vector RO, pointing from R towards 0. 

If the planes of the given couples have no common perpendicular 




Fig. 16 



THE COMBINATION OF A COUPLE AND A SINGLE FORCE 



21 




plane, the polygon OPQSTR is not a plane figure; it is a polygon 
in space, but OR is still the vector of the resultant couple. 

33. The combination of a couple and a single force acting in a 
plane parallel to the plane of the couple. 

Suppose that among the forces which act upon a body at rest there 
are two which form a couple and a third force whose line of action 
is in, or parallel to, the plane of the couple. Let the irregular outline 
(Fig. 17) represent the stationary body; in this case, 
let the curved arrow and the numerical value of M 
represent the moment of the couple; and let the 
forced be represented by OA. It will now be shown 
that the resultant of M and F is the single force F r . 

(a) If M and F are not in the same plane, shift M Fig - 17 
to a parallel plane containing F. 

(b) In place of the original force and arm of the couple, let M = Fl, 
in which F has the magnitude of the given force F, and I = M/F. 

(c) Shift and turn the couple till one of its forces F', has the line 
of action of the given force F, but acting opposite to it, thereby can- 
celing it, so that nothing is left but the force F r acting parallel to the 
given F, and of the same magnitude. The distance moved to the left is 
l = M/F; that is, if an observer standing at sees the couple right- 
handed as he looks down, the Resultant of M and F, will be F r on his 
left, when he wheels about and looks in the direction OA. 

Problem. The student should show that a given left-handed couple 
would have caused the resultant force to appear on his right. 

34. The converse of the above proposition is evidently true, viz.: 

A given force F may be resolved into a couple and an equal force F 

acting at any point in space, distant I from the line of action of F, 

the moment of the couple being Fl and its axis being perpendicular 

to the plane of F and I. rt ^ ml 

H 35. The resolution of 

couples. A given couple 
may be resolved into two 
or more couples graphic- 
ally, if the vector of the 
given couple is the line AB 
(Fig. 18 b), by drawing two 
(&) K component vectors AH and 

HB; or by three vectors 
which may be the edges of a rectangular solid. (Fig. 18 (a) ). 







/ 






B 

M s 


/ 


4 


/m 




■xz 



22 THE THEORY OF COUPLES 

30. Resolution of a force into a force and two couples. A 

single force acting on a body, may be resolved into two couples and 

an equal single force acting at any point on the given body that may 

be chosen. 

Let the point A (Fig. 19) be the point of action of a given force F. It 

may be resolved by the introduction 
of two balancing forces +F and — F, 
at B, into an equal parallel force at 
B, and a couple Fy> right-handed as 
seen from -\-X 9 as shown in the figure. 
Still more, the force +F at B may 
be itself resolved into an equal paral- 
lel force +FatO, and a couple ( — Fx) 
(which is left-handed seen from +F). 
We have thus resolved a force, acting 
at the point (.r, y) and parallel to 
0Z y into an equal force acting at 0, 

and two couples; one having a moment -\-Fy about the X-axis, 

and one having a moment —Fx about the F-axis. 

The use of this method of resolution will be illustrated by several 
examples in Chapter III, and will be freely employed in subsequent 
chapters. 




CHAPTER III. 



Parallel Forces Which Balance. 

37. Three parallel forces balance. Since they could not balance, 
if all acted in the same direction, the largest must act in a direction 
opposite to that of the other two; hence we must have 



or, numerically, 



F 1 +F 2 -F 3 = 
F3 = Fi+F 2 . 



(i) 



If we resolve F 3 into two forces respectively equal to F\ and F 2 , 
we have two couples which must balance each other. Since they 
could not balance if they were not co-axial, g,nd since the couples 
cannot be co-axial unless the forces are co-planar, we reach one im- 



PARALLEL FORCES WHICH BALANCE 



23- 



portant conclusion : that three parallel forces cannot balance unless 
they act in the same plane. 

Since the couples balance, their moments 
have equal magnitudes and opposite signs. 

Ex. Let the three forces be represented by 
a scale drawing, Fig. 20. As we look down 
upon the drawing, Fib is right-handed, and 
F 2 a is left-handed. Hence Fib — F 2 a = 0, 




Fig. 20 



a 



Fi 

or — = - 
F 2 b 



(2) 



or the forces (of the balancing couples) are inversely as their arms. 

If we let the distance between the forces having the same direction 
be known as c, we have in this case 

c = a-\-b 

Now erect an axis at A, and take moments. We have at once 

Fzb = F 2 c, hence — = - 

^3 c 

If we take C as our axis, and take moments, we have 



(3) 



FiC = Fsd hence — = - 
F* c 



(4) 



c; 



(5) 



Combining (2), (3), and (4) we have 

Fi : F 2 : F 3 = a : b 
a very useful triple proportion. 

38. Deductions. From the above discussion several important 
conclusions follow. When three parallel forces acting on the same 
rigid body balance, we know: 

1. The three lines of action lie in the same plane. 

2. Their algebraic sum is zero. 

3. The line of action of the largest force lies between the lines of 
action of the other two. 

4. The magnitude of each force is proportional to the distance 
between the other two. 

5. In the language of algebra 



Fi+F 2 +F 3 = 

Fi : F 2 : F 3 = a 

a-\-b = c 



b : c 



(6) 

W T hen a body acted upon by three parallel forces is at rest, we know 
that the above equations must be true. 



CO-PLANAR PARALLEL FORCES 



Problems. 

1. Given two parallel forces 18 lb and 40 lb, 8 feet apart. Find 
the balancing force, and the distances between it and them. 

2. Given the parallel forces P=+80 lb, Q=-64> lb, and the 
distance between them, 12 feet. Find the balancing force, and its 
position. 

3. Three parallel co-planer forces act on the same straight bar at 

given points A, B, and C, and 
T balance. The force at A = 35 lb. 

T "P* Find the forces at C and B. 

Use the proportions in equa- 
tion (6) for magnitudes. 

4. Suppose the force A makes an angle of 30° with the bar. 
How does the obliquity affect 

the result? v^ .„ \ b *^^0# 



Fig. 21 



Fig. 22 



39. The resultant of two 
parallel forces. Since F u F 2 , 

and F 3 balance, it is evident that one must balance the combined action 
of the other two. Now the single force which is equivalent to the com- 
bined action of two or more forces is called the "Resultant" of those 
forces. In the present case the resultant of two must be equal and 
directly opposite to the balancing force*. Hence if we know all 
about the third or balancing force, we know all about the 
resultant. 

Thus, if P, Q and S, acting on the same body, balance, the resultant 
of P and Q is directly opposite to S; and the resultant 
p of P and S is directly opposite to Q. 

The student should adapt the equation (6) to this 
?! j, ^ case; and write deductions as to the resultant as was 
done for the balancing force in the last section. 

40. Graphical solution for the resultant of two 
^^ * parallel forces. The following elegant solution is 
taken from Rankine's Applied Mechanics. 

Ex. Given, in Fig. 24, two parallel forces, F\ and 
F 2 , fully represented in magnitude, relative position, and direction by 
AP and BQ. To find their resultant graphically. 

Solution. 1. Draw thru P and B lines parallel to the (imaginary) 
line AQ. 2. Draw thru A and Q lines parallel to the (imaginary) 



Fig. 23 



* This justifies Rankine's definition: "The resultant of a system of forces is the 
force which balances the single force which balances the given forces." 



GRAPHICAL SOLUTION 



25 



line PB. The intersection of the new lines thru P and Q will give the 
arrow head of the resultant; and the two lines thru A and B will 
intersect at the tail end of the resultant, which is therefore CR. 

In (a) Fig. 24, the 
given forces act in the 
same direction: in (6), 
they act in opposite 
directions. 

The geometrical proof 
of the correctness 
of these solutions is left 
to the student. He 
should prove in each 

case that CR fulfills all requirements of 
position. 




magnitude, direction and 



41. The lever. Case I. Suppose a rigid beam carrying a hook 
at each end is balanced on a knife edge, i. e. y the sum of the pulls of 
the earth on the beam and hooks is equivalent to a single pull directly 
over the knife edge. This pull (or weight) is balanced by the up- 
ward lift of the knife edge, or ''Fulcrum/ 9 so that when we come to con- 
sider the action of other forces on the beam, we can completely neglect 
the weight of the beam itself. 

Let us represent, Fig. 25, the distance from hook to hook by Z, and 
the length of the two arms by a and b respectively. 

Suppose Fi and F 2 are applied to the hooks and that they balance. 

We have at once D „ . „ 

ti = r i-\-t 2 



which must be the additional 
support given by the fulcrum 
— to balance Fi and F 2 . 

From the proportion equation 
derived from (6) 




Fig. 25 



l = F 1 :F 2 :R 



FJ 
R 



FJ 



Fi+Ft 



, and b = 



F 2 l 
R 



Fd 



F,+F 2 



Again, if F\, F 2i and a are given, we have as before, R = Fi~\-F^ and 

from the proportion ^ 

r i a 

FT" b' 



and l = a-\-b. 



b= — a. 



26 THE THEORY OF LEVERS 

If Fi and the two lengths, a and 6, are known 

then F 2 = F\ 
a 

a + b 

and R= 1 i. 

a 

This case was formerly spoken of as a "lever of the first class.'' 

42. Case 2. Let the lever AB and its hooks be balanced on C 
by means of a counter weight D, so that the weight of lever and counter- 
weight may again be neglected. See Fig. 26. Let F 2 be a known 

force acting down as shown at a distance 
a from C. An unknown force acts up at 



f d < — c — J> i? whose distance from F 2 is c. We are to 



A t^^ir* 



/^— a find the force Fi and the additional load put 

fg upon the support at C. 

It is evident at a glance that the turn- 
Fig. 26 |^J j n g momen t of Fi about a horizontal axis 
at C is negative (left-handed), and equal to 
Fi(a-f-c). It is also evident that the turning moment of F 2 about 
the same axis is right-handed, and equal to F 2 a. 
These two moments must balance, hence 

Fi(a + c) =F 2 a 
so that 

F t -JL ft. 

But a+c = 6 as defined for the lever, so that we have 

— = — as in formula (6). 
F 2 b 

The resultant of F x and F 2 must be equal to their difference and it 
must act opposite to the support of the fulcrum, as the latter is the 
balancing force. Hence the fulcrum action due to Fi and F 2 is 

R = F 2 -F 1 

Fig. 26 illustrates what was once known as a "lever of the second 
class." 

43. Case III. A lever of the third class is illustrated by Fig. 27. 

Ex. 1. A man, with a pitchfork six feet long, supports at the far 
end a mass of hay weighing 24 lb. One hand is at A, acting down, 
and the other hand at B, 18 inches away, is acting up. What are the 



FRICTION VARIES FROM ZERO TO A LIMIT 



27 



forces at A and B if the handle is horizontal and all actions are 
vertical? 

The hand at A may be thought of as the fulcrum. 

Ex. 2. In the above example, for pitchfork substitute fish pole; 
for 6 feet, put 20 feet, and for hayweighing 24 lb, put a fish weigh- 
ing 5 lb. Find the forces at A and 
B, capable of just supporting the 
weight of the fish. 



t-l-.B- 






Fig. 27 



44. A Word about Friction. 

Whenever the surface of one body 

is in contact with the surface of 

another, there is always a more or 

less effective interlocking of the 

minute particles of one body with 

the particles of the other, so that they are able to resist separation by 

sliding. This ability to resist sliding (or relative motion) is variously 

called adhesion, cohesion, sticking, or friction. The term Friction is 

used in Mechanics as the name of the force with which the bodies in 

contact resist a mutual sliding. 

There is, of course, a limit to the ability of material in contact to 

resist a sliding motion. When the external forces tending to produce 

sliding are so great that the limit of frictional resistance is passed, 

then sliding takes place. 

Fig. 28 represents a "rough" heavy body resting upon a 

horizontal plane which is not smooth. The body is stationary in 

spite of the tension T in a rope which 
tends to draw it sideways. As the body 
A stands still, the slender blocks (6) stand 
on end in equilibrium. The friction be- 
tween A and B is sufficient to balance 
T. Between A and B there are two 
different actions both distributed over 
the surface of contact, one normal or 
vertical, one tangential or horizontal. 
Each of these forces is represented by 

two arrows. The two arrows on A show how A is acting on B; the 

two arrows on B show how B is acting on A. It is evident that 5's 

horizontal action on A is equal to the horizontal action of the rope T, 

and in the opposite direction; hence they balance. 

Let F denote the friction shown by the horizontal arrow on A or B. 

Since A stands still we have 

F=T 




Fig. 28 



28 THE THEORY OF LEVERS 

If more shot or pebbles are put into C, T will be increased, and of 
course F will be increased also, as long as A stands still. When T 
increases to a certain point, A starts and slides. F has reached its 
limiting value F u and the difference between T and Fi makes A move 
and perhaps causes the slender blocks to fall over. 

In statical problems the limit of frictional ability is not reached. The 
laws which govern friction will be discussed later when motion is 
studied, but it is well, at this point, to remark that the frictional 
ability of surfaces in contact depends upon several things: the normal 
action (pressure) between the two; the character of the surfaces 
(rough, polished, or lubricated), the temperature, etc. In Mechanics, 
the word "smooth" is used only in cases of ideal bodies where the 
friction is zero. 

45. Applications. Case I. Show how the action of steam on 
the piston of a locomotive engine has a tendency to move the loco- 




yy^yyy 



Fig. 29 



motive; -first, when the crank pin is above the axle of the driving wheel 

(Fig. 29). 

Solution, 1. Suppose the steam at high pressure is as shown in the 
cylinder. It presses on the piston towards the right and equally on 
the cylinder head towards the left. The cylinder is bolted to the 
frame, so the action of the steam on cylinder head is to push or move 
the frame of the engine to the left. If the area of the piston face is 
A square inches, and the steam pressure per square inch is p lbs., 
in excess of the "back pressure" on the other face, the force tending 
to move the frame to the left, or backwards, is Ap. 

The steam pressure, Ap, upon the piston is transmitted thru 
the piston rod b to the crosshead B> thence thru the connecting rod c 
to the crank-pin Q. Omitting for the present the small amount of 
frictional resistance offered: — by the cylinder upon the circumference 




HOW LEVERS MOVE THE LOCOMOTIVE 29 

of the piston, by the packing in the cylinder-head upon the piston- 
rod; and by an upper guide D upon the cross-head, — it is clear that the 
horizontal force acting to the right upon the crank-pin at Q is Ap. 

The crank and wheel form a rigid lever which (omitting the pull of 
the earth which is balanced by the vertical action of the track) is 
acted upon by three forces which balance. (See Fig. 30.) 

First, The force Ap on the crank pin; 

Second, The frictional force of the track upon the circumference of 
the wheel at E, which force we will call T. This force prevents slipping. 

Third, The reaction of the frame thru 
the journal box at C upon the axle, which 
reaction must equal the sum Ap-\- T. Thus 
we see that the direct action of the axle 
upon the frame is the resultant of Ap and 
I 7 , which is Ap-\- T acting towards the right. 
We found in the first paragraph of this 
solution that the action of the steam upon 
the cylinder-head tended to move the 
engine frame of the locomotive to the left, 
by a force of Ap pounds. We have in 

the last sentence found that the action of the axle of the driving wheel 
tended to move the frame to the right by a force of Ap-\-T pounds. 
The algebraic sum of these horizontal forces is 

Ap+T-Ap=T 

which is exactly the horizontal action or friction between the track 

and the wheel. 

We shall find the numerical value of T by solving the problem of 

the lever consisting of the radius R and the crank r. By the law of 

the lever . „ . 

A P _ g or T _ rAp 

T r R 

which is the measure of the resultant forward action of the steam in 
our cylinder upon the locomotive; in other words, this is the force 
which tends to move the locomotive, and which will move it unless 
some force not considered above holds it. The student must not 
fail to see clearly that the action T of the track upon the wheel is forward, 
and hence the locomotive moves forward. 

46. Second, when the crank is below the axle. Case II. See 

Fig. 31. 

1. In this case the action of the steam upon the cylinder head and 
thru it upon the frame of the locomotive, is Ap to the right. 



30 



THE THEORY OF LEVERS 



2. The action of the steam upon the piston puts compression or 
thrust into the piston-rod and into the connecting-rod, and has a 
tendency to move the crank-pin towards the left. 

3. As before, the rigid body of wheel and crank forms a lever, 





Fig. 31 



with two parallel forces, Ap and T acting in opposite directions. Fig. 32. 
Their resultant acts towards the left thru the axle upon the frame, 
namely, Ap— T. (See the lever of the third class), 42. 

4. The sum of the actions upon the frame* is Ap on the cylinder- 
head, which is positive to the right, and Ap— T 
which is negative to the left. Their resultant 

Action is : . T . T 

Ap-(Ap-T) = T, 

and T is jnst the same as before. This shows 
that the steam is urging the locomotive forward 
just as much when the piston rod is pushing 
back, as when it is pulling forward. 

In the foregoing discussion it is very import- 
ant to keep in mind the vast difference between 
the action of the Frame upon the driving axle 
(thereby making a balancing force with T and Ap), and the action of 
the axle upon the frame thereby being the resultant of T and Ap. It 
is this resultant which counts in the discussion. 




47. Remarks. From the value oi T = ^^ i n each case, it 
appears : — ^ 

1. That the higher the steam pressure p, the greater the traction. 

2. The larger the cylinder cross-section, the greater the traction. 

3. The longer the crank-arm, the greater the traction. 

4. The smaller the radius of the wheel, the greater the traction. 

5. Since T is limited by the frictional ability of the surfaces in 
contact (wheel and rail) which ability depends, as everybody knows, 



HOW TRACTION IS INCREASED 



31 



upon the load on the wheel, and the character of the surfaces (clean 
or sanded, wet or dry, cold or hot, greasy or icy, etc.) there is in every 
case a greatest possible value of T, which we will call T\. 

Since the value of T\ is made greater by increasing the roughness 
or "biting" ability of the rail, the engineer, when he needs greater 
traction, operates a device which sprinkles sand upon the rail. But 
even then there is a limit to the value of 2\. 
Hence the greatest value of the steam pressure is 

RT l 

Vl = — — 
Ar 

If the engineer attempts to increase 2\ by increasing p u the wheel 
will slip, and as is well known the traction at once falls below T\. 
However, the traction can be increased by the addition of one or two 
driving wheels, each a duplicate of the first and each carrying, approxi- 
mately, the same load. The crank or cranks must be similarly placed, 
and the new connecting-rod, or rods, from wheel to wheel, must 
always be horizontal. The action of the main connecting rod is then 
distributed between the wheels, thus permitting a greater steam 
pressure without causing the wheels to slip. T is then %T. 

Problem. 

Let the student make skeleton drawings of a horizontal engine 
driving a paddle-wheel steamboat, and then write out analyses similar 
to the above.* 

The problem in the text is introduced to show that the principles 
of the lever enable one to explain how 
steam drives a locomotive. The formula 
found holds for a single cylinder and 
only when the crank pin is directly above 
or below the axle, 

46. Bent Levers. Bent levers are 
analyzed as are straight levers by the use 
of moments. A lever (without weight) 
acted upon by forces Fi and F 2 , with 
lever arms n and r 2 respectively, is in 
equilibrium if the moments Fin and F 2 r 2 are equal, provided the axle 
at C is properly supported. The angle ACB may have any value. 




* The captain of a Mississippi steamboat once tried to convince the author that 
his engine propelled the boat only when it was pulling forward with the crank pin 
above the shaft. When the crank was below the axle and the steam was pushing 
back, he thought the engine was a hindrance and not a help. He knew so little of 
the principles of Applied Mechanics that no reasoning could convince him that he 
was wrong; but his boat went ahead with every stroke just the same. 



32 COMPOUND LEVERS 

F r 

Hence — = — or the forces are inversely as the arms, as in the case 

F 2 n 
of a straight lever. 

The Support reactions. The reactions upon the fixed axis of 
a bent lever must not be overlooked. In the last figure the force 
Fi calls into being the equal acting force Fi' = Fi; and the force F 2 
in like manner brings into play the action F 2 = F 2 . (See 19.) The 
resultant of F\ and F 2 (see Chapter V.) is the force 

R = V 0? + F 2 2 - 2FiF 2 cos^t CB) , 

which must be balanced by the reaction of the journal box of the axle. 
The magnitude and direction of this supporting force must never 
be ignored. 

47. Compound levers. The chief use of levers is to transmit 
and modify force. A force is applied to a lever at A, and the result 
is a very different force applied to a body at B. A very small 
force at A may produce a very large force at B, given a strong lever 
and a sufficient support as a fulcrum, and there is no limit to the 
ratio of the force at B to that at A. Hence, the first great Engineer 
of history, Archimedes, said: — "Give me a spot whereon to rest my 
lever [that is, a fulcrum] and I will move the world." 

1. It is often necessary or at least profitable to separate a ratio 
into the product of two or more ratios, which means that one lever 
may be made to operate a second lever, and, if needed, the second 
may operate a third. Thus: Fig. 34. 

It is evident that 

F* 24 
F 2 = 4Fi; F 3 = 6F 2 = UF i; F 4 = — = — Fi. 

7 7 



-5fc--l'-: 



IT 



2_ 



3 



A Compound Lever. Fig. 34 



'-f- 



The first lever is of the "First Class"; the next of the "Second 
Class," and the last is of the "Third Class." In every analysis the 
fulcrum should be taken as the moment axis. A compound lever 
may be designed to make the ratio F 4 large or small. In Fig. 34 
the levers must be so proportioned as to balance when F\ is zero, so 
that their weight may be neglected without error. The same will 
be true of those examples which follow unless otherwise accounted for. 



WHEELS AND PULLEYS ARE ROUND LEVERS 

2. Fig. 35 is equivalent to a compound lever consisting 
simple levers. The student will readily see that, assuming 
tion and no stiffness in any rope (a very 
violent assumption): 

HL -^ Sl . * . * 

Fi n r 2 n 



of three 
no fric- 



That is, the action of W is exactly balanced 
by the action of F\ 9 thru the intervening cords 
and pulleys. If there be friction at the axles, 
or stiffness in the cords, then the weight W 
will remain in equilibrium in spite of some 
departure from the value given by the formula: 

Fl = r 2JlS*W. ■ 

R1R2RZ 

In Fig. 35, all pulleys and drums are mounted on fixed supports. 
In the same way gear wheels with their axles act as levers in a train 
of wheel work. Combinations of pulleys with fixed and with movable 
blocks will be discussed later in the book. 




Fig. 35 



Problem. 

48. Pullman Car Truck. In an ordinary Pullman car, one- 
fourth of the weight of the car body and contents is brought to each 
set of three wheels on one side of a truck. It is then by a system of 
levers distributed in such a way that each wheel carries the same 
load, viz.: one-twelfth of the entire weight of the car.* The student 
should inspect a truck, and by means of a drawing explain how this 
is done. 

Testing Machine of St. Louis Bridge. 

49. A fine example of compound levers is found in the testing 
machine designed by Col. Henry Flad and used by Capt. James B. 
Eads in testing the strength and elasticity of the materials which 
were put into the great Eads Bridge at St. Louis. The arrangement 
of the levers is shown in the "figured" drawing, Fig. 36. 

This machine was capable of exerting a tensile or compressive 
force of 100 tons, by means of fluid pressure in a thick iron cylinder, 
and a system of compound levers. 



*A car "truck" is the low steel carriage with six wheels upon which one end 
of the car-body rests. 



34 THE ST. LOUIS TESTING MACHINE 

When a specimen of steel was to be tested in tension it was coupled 
in between the two pins shown in the horizontal section near the 
figure (6") in Fig. 3. The draw-bar, or piston-rod, is of steel 3 inches 
in diameter; it passes into the cylinder and is screwed into the plunger 
or piston. Liquid, glycerine and water, was forced into the cylinder 
by a force pump operated by hand. The plunger was 7j inches 
in diameter, and a liquid pressure of 6,000 lbs. per square inch gave 
a total of 100 tons. The cylinder itself was held in place by large 
lugs between blocks on the frame of the machine. 

When a specimen was to be tested in compression it was inserted 
between the parallel steel faces on either side of the letter E. When 
the plunger was forced out it pushed the specimen against the outer 
cross-head (on the left) which brought the four long tension rods 
into action; they transferred their joint action to the short arm of 
the first lever, as was done directly in the former case. 

The extreme delicacy of the apparatus is shown by the fact, reported 
by Capt. Eads, that when a specimen was under a stress of nearly 100 
tons, a common lead pencil dropt into the can at K would cause the 
long arm of the last lever to fall; and upon the removal of the pencil 
the arm would rise again. Every lever turns on a steel " knife-edge' * 
which acts as a fulcrum. 

The first lever is a bent lever of the first class. Its short arm is 
10j inches; its long arm is 9 feet 4f inches; so that the ratio is 1 to 11. 

The second lever is of the third class. The short arm is 12 inches; 
and the long arm is 13 feet. Hence the ratio is 1 to 13. 

The third lever is a simple straight lever of the first class. Its 
short arm is 7 inches, and the long arm is 8 feet 1.86 inches, so that 
the ratio is 1 to 13.98. 

All the levers were balanced so that the third lever, which should 
oscillate between very narrow limits, stood horizontally when there 
was neither tension nor compression in the specimen. A suspended 
pan shown at the end of the third lever was in great part the balanc- 
ing force for all levers. 

When the liquid used to drive the piston or plunger in the thick 
cylinder was applied, the action began upon the specimen, which 
had been inserted into the machine. The balance of the levers was 
maintained by pouring fine shot into the pan, and the weight W, of 
the shot, having been carefully weighed in an accurate balance (such 
as is used in a chemical laboratory) furnished the means for measuring 
the tensile and compressive stress applied to the specimen. The 
formula for such a calculation was: 

If P denotes the required total stress 

P = 11 X 13 X 13.98X^ = 1999.2 w = 2000 w, very nearly. 



WARNING TO SELF-STYLED INVENTORS 



35 



If w was 100 pounds, P was 100 tons, which was regarded as all 
that it was prudent to apply to the machine.* 

Suppose a steel cylinder 8/10 of an inch in diameter is under ten- 
sion, and that the shot dropped into the can, to produce a balance, 
weighs 8 lbs. and 3| ounces. What is the tension per square inch of 
the cross-section in the specimen? 

47. Another warning. The student who has seen, in what has been 
shown, how a small force at one point can by means of levers pro- 
duce or balance a large force at another point, must not jump to the 
conclusion (as so many self-styled inventors have done) that the 
increase in the force thus produced is all clear gain. Levers are of 
practical use, as abundant experience shows, but they do not enable 
one to get something out of nothing. 

Levers are generally used to move things, to lift heavy loads, to 
do work. We shall discuss motion later on, but it is timely to say 
now, that when motion results, there is full compensation for any 
apparent gain in force by a loss of motion; and a balancing loss in 
force for any gain in distance moved. 

Ex. For example, if a force of 20 lbs. (see 
Fig. 37) can balance a weight of 1,000 lbs., 
which is 50 times as great, the body pulling 
down on the rope at P must pull the rope 
down 50 feet in order to raise W one foot. 

See chapter XXI on Work and Energy. 

48. The Relation between any number 
of Parallel Co- Planar Forces acting upon a 
body in equilibrium. Since the forces balance, 
there can be no resultant tendency to move the 
body in any direction or to turn it around. 
Hence, the algebraic sum of the forces must 
be zero; and the sum of their moments with 
reference to an axis erected at any point in 

their plane must also be zero. Hence, the Equations of Equilibrium. 

2M=0 




Pig. 37 



* This elegant machine is still in use in a St. Louis Manufacturing establish- 
ment. Figure 36 is reduced from a lithographic plate in the author's "History of 
the St. Louis Bridge. ' ' 

For further description of the machine, and the elegant method employed for 
determining elongation, modulus of Elasticity, and Elastic limit, devised by Chan- 
cellor Chauvenet of Washington University, the reader is referred to the History, 
published in 1881. 



36 



PARALLEL FORCES IN A PLANE 



In point of fact the balancing forces can always be resolved into 
a number of separate couples which balance. 

1. Example. Let the forces be Fi F 4 in Fig. 38. The 

force F 3 = 16 can be resolved into 8+4+4, 
which, taken with /\ = 8, F 2 = 4, and F 4 = 4, 
form three couples whose algebraic sum is 
evidently zero, just as 2Af = for any axis 
as 0. 

2. Example. Six parallel co-planar forces 

as shown in Fig. 39. What and where is the 

Take the line AB as the axis 



t 



Fig. 38 






2' ><i': 



Fig. 39 



10 



act upon the bar AB, 
single force which will balance them? 
of X, and any point on it as 0, as an 
origin. Let the unknown force be P and 
its distance to the right of be X. Sub- 
stitute in the Eq's of Equilibrium, and 
solve. Let forces acting down be posi- 
tive so that (+F) (+x) = +Fx a right- 
hand moment. 

3. Ex. Suppose we have a rigid frame ABCD, lying on a 
smooth horizontal plane (or floating on the horizontal surface of 
water), so that its weight is balanced and left out of account. Suppose 
further that the frame is acted upon by a set of parallel forces all 
in a horizontal plane, and that they balance. To make the situation 
seem concrete, let every arrow in Fig. 21 represent the tension in a 
wire which is fastened to a pin in the frame. Every wire is connected 
with a spring balance, which does not appear in the drawing. All 
dimensions are known, and all the forces are given with the exception 
of two: P and Q. These two are to be found. 

Solution. Since the forces balance, there can be no resultant force 
in any direction, and no resultant moment. Hence we must have 
2F = and Hif = 0; in which summations both P and Q are included. 

If the moment axis is taken 
in the line of action of one of 
the unknown forces, that force 
will not appear in the moment 
equation, since a force cannot 
have a moment about an axis 
which it intersects. Hence, at 
first take the axis at A, 
Let the triangles of the frames be equilateral with a side 16 feet long. 

Let Fi = 3501b F 3 = 1001b F 5 = 2001b 

F 2 = 6001b F A = - 7401b F 6 = 4001b 



p 


B 


T^ 


C Q 


v 


o 


la a\ / 








J 


l< "" 16 '"' Hk 


?» [ F 3 [ F 5 


?6 


D 



Fig. 40 



SUPPORTS OF A BRIDGE TRUSS 



37 



The equations of equilibrium then become 

350 + 600 + 100 -740+400-P-§ = 
or P + Q = 910; and 

16 X350 + 24X600+32X 100-40X740+48 X200+56X400-64Q = 0. 

It is at once evident that this last equation is clumsy; a "half -panel" 
length could have been used as a unit of length instead of the foot, 
so that the "arms" would have been 2, 3, 4, 6, etc. Thus: 

2 X350 + 3X600 + 4X 100-5 X 740 + 6 X200 + 7X400-8Q = 

or Q = 4001b 
and therefore P = 5101b 

Let the student take his axis at D (thereby eliminating Q) and re- 
calculate the value of P. 

The student must not suppose that the equation 2(F) =0, 
and two moment equations, arising from the use of two different 
axes, furnish three independent equations sufficient for finding three 
unknown forces. In fact, the moment equations are not independent. 

49. Again, suppose the frame is a bridge truss (Fig. 4 1) in a vertical 
plane carrying half of a loaded roadway. Let it be assumed that all 
forces are vertical and co- 
planar, and that the loads [ 2 T 
are placed at the joints. 
Assume any convenient 
dimension for the sides of 
the triangles which are 
equilateral. Find the sup- 
ports at A and B. It will 
be well to let every bar in 

the frame have the length 
£. It will be seen that the 
absolute length has no 
effect upon the values of 
Fi and V 2 . 

Ex. 1. Figure ABC 
(Fig. 42) is a cantilever 
truss, designed to support 
a heavy load at C; and 
smaller loads at the upper 
pins. It is assumed to be 
rigid in a vertical plane. 




1, 1' 


v l 

1 


u 


't 


•> 

10 

Fig. 


T 
41 


18 


T 


% 


1 J 












38 



PARALLEL FORCES IN A PLANE 



Let Fi = 8001b F 3 = 5001b F 2 = l,000lb JF=l,600lb 

Find the supporting forces at A and 5. 



Ex. 2. Let 



Fi = 1,2001b F 3 = 200lb 
F 2 = 1,0001b PF = 60lb 



Find the supporting forces at ^4 and B. 

50. To find the resultant of a set of parallel co-planer forces 
which do not balance. The magnitude of R is evidently equal to 
the algebraic sum of the given forces. Hence R = HF. Its position 
is found from the necessary fact that the moment of R about any 
axis is equal to the algebraic sum of the moments of its components 
about the same axis; that is: Rx r = H(Fx). 



or x a 



SF.t 
R 



R 



This is really the same as was shown in 18, when we combined 
a force and a couple. Every moment comes from a couple, and in all 
statical problems that couple is always easily found. 

Ex. 1. Suppose a floating platform is to be acted upon by a system 
of parallel forces in a horizontal plane which do not balance, and that 
we are to find the balancing force. 

In order to find the required balancing force, or balancing couple, 
find the resultant; and then balance it. Let the platform, Fig. 43, 

2i be rectangular, and let the 
given forces all act parallel 
to the edge OY, and let 
their distances from Y be 
known The data and cal- 
culation may be tabulated 
as shown in the table. Each 
force is resolved into a force 
rig. 43 ** along Y and a couple, as 

explained in 19. 
The forces at are then added, and the moments about a vertical 
axis OZ, are also added. 

So that R = HF 

and M = Z(Fx) 






FINDING THE BALANCING FORCE 



F 


X 


r* 


+ 12 


3 


+ 36 


-10 


10 


-100 


+ 48 


15 


+ 720 


-24 


22 


-528 


# = 26 




M = 128 



The resultant force along OF is +26, and the resultant moment 
is +128. The resultant force and the resultant couple combined 
(see 18) give a single force +26 at a distance from equal to 
M/R = 4tii- Hence the single resultant of 
the given forces is a force of 26 lbs. act- 
ing in a positive direction and at a dis- 
tance x r = 4x1 • It follows that the balanc- 
ing force is 26 fibs, and that it acts in the 
negative direction along the line of R. If 
this force, F 5 , were added to the given 
forces, the table would show complete 
equilibrium, since we should have R = 
and M = 0. 

Ex. 2. Find the resultant of the follow- 
ing system of forces applied to a similar body with opportunity for 
negative values of x. It will be found that the resultant is a couple, 
and that only a couple can balance the given forces. See Table. 

51. Parallel forces in space. It is easy to con- 
ceive a rigid body acted upon by a variety of other 
bodies along parallel lines, with concentrated forces 
and lines of action so adjusted and proportioned 
that the body is at rest. In fact, every car that 
stands on the track, every house that stands on 
posts or columns, every table that stands on legs, 
every loaded wharf that stands on piles, furnishes 
an illustration of this proposition. 
Since the forces balance, there must be no resultant tendency to 

move the body in any direction. This gives us one equation between 

the magnitudes of the forces: 



F 


X 


M 


+ 40 
-16 


+ 4 
+ 6 

— 2 






-36 




+ 12 


-5 



2F = F!+F 2 + 



F =0 



In like manner, since there can be no resultant tendency to turn 
the body about any axis in space, we have another general equation 
involving both magnitudes and directions: JIM = 0, the axis of M 
being any line in space; more especially a line perpendicular to the 
given forces. 

It was shown in 35 that a moment about any axis in space could 
be resolved into three component moments about three rectangular 
axes; that is , , 

M=V (Mi 2 +if 2 2 +M 3 2 ) 



if the axes referred to are OX, OY and OZ. These axes may be taken 



40 



PARALLEL FORCES IN SPACE 



at will. Let OZ be parallel to the common direction of the lines of 
action of all the forces. It follows that 

M 3 = 

inasmuch as a force can have no direct tendency to turn a body about 
a line parallel to its own line of action. There remains therefore 
M = V (Mi 2 -\-M 2 2 .) Now, M cannot be zero, unless both Mi and M 2 
are zero. Hence the "Equations of Equilibrium" are 



2(F) =0) 

l/ 2 = oJ 

52. The choice of co-ordinate axes. In accord with what 
appears to be the best usage, in all space problems in this book, 
the axis OZ will generally be taken vertical; the axes OX and OY are 
so taken that positive rotation about OX looking from -\-X towards 
carries + Y into -\-Z; and a positive rotation around OY carries 
+Z at once into -\-X. It is well to note the sequence of letters 
and numbers: X> F, Z; 1, 2, 3. Rectangular axes will accordingly 
stand thus: (Fig. 44). All the co-ordinates of P are shown positive. 
Arrows on the axes show how one must look at a couple, or moment. 

An examination of Fig. 44 will disclose 
the fact (shown again for emphasis) that 
Fy is a right handed moment to an observer 
looking from +X towards 0, whenever F 
and y are both positive, or both negative; 
and that the moment is left-handed or 
negative whenever the factors have differ- 
ent signs. Accordingly, the algebraic law 
of signs holds for every term in H(Fy). 

It is just the reverse for every term in the expression 21 (Fx). When 
the algebraic product is positive, the moment is left-handed and 
therefore negative; and when the algebraic product is negative the 
moment is positive. 

Hence Jfi = 2(Fy) \ 

M 2 =-2(Fx) J 

This apparent anomaly of a negative product from positive factors 
is unavoidable. 

Ex. Take as an illustration an ideal system of balanced forces 
which are parallel and may be tabulated as follows: 



Hri- 




Fig. 44 



POSITIVE AND NEGATIVE MOMENTS 



41 



The forces represented 
in the table are fairly 
shown in this space draw- 
ing, Fig. 45. The points 
of application are assumed 
to be in plane XY. The 
student may combine 
these forces by pairs 
according to any method 
already given, and then 
combine the partial result- 
ant until he finds remain- 
ing two resultant forces 
which will directly balance 
each other. This checking process should not be neglected. 



F 


X 


y 


Fx 


Fy 


+ 8 


+ 2 


+ 3 


+ 16 


+ 24 


-6 


+ 3 


-1 


-18 


+ 6 


+ 4 


-1 





- 4 





-3 


+ 2 


+5 


- 6 


-15 


+ 2 


+ 1 


-5 


+ 2 


-10 


-5 


-2 


+ 1 


+ 10 


- 5 


2F = 






l>(Fx) = 


2(*V)= 



The 




+x 



Fig. 45 



forces shown should be laid off with great accuracy. 

53. The equations of equilibrium. 

It is readily understood that one may 
know that a system of parallel forces 
balances, yet be ignorant of the numer- 
ical values of some of the quantities which 
enter into the equations of equilibrium. 
The three equations of (51) will suffice 
to find three unknown quantities, pro- 
vided every equation has at least one unknown quantity in it. 

There are three general cases each of which will be illustrated. 
The three unknowns may be: — 

1. Three unknown forces. 

2. Two unknown forces and one co-ordinate. 

3. One unknown force and two co-ordinates — one an x, and the 
other a y. 

Case I. Three forces are unknown. Suppose a rigid slab, 
rectangular in shape, is supported in a horizontal position by four 
vertical wires connected with adjustable pins, and that one of the 
wires includes a spring balance which shows the tension; the other 
three tensions are to be found. 
All the points of attachment 
with the slab are known. 

Let the length of the slab 
be twelve (12) feet, and the 
width four (4) feet, and as- 
sume that the slab weighs Pi g . 46 




42 



PARALLEL FORCES IN SPACE 



960 lbs. The axes assumed, and the position of the four supporting 
forces are given in Fig. 46. The known force is F. The three equa- 



tions of equilibrium are 



P + Q+r+24-960 = 

M 1 = 24 X2 + PX1 + TX4 -960X2 = 
M 2 = 960X6 -24 X12-PX9-QX6- TX3 = 



(1) 
(2) 

(3) 



Eliminate Q from (1) and (3) and get (4). 

Eliminate P from (2) and (4) and get the value of T, and so on. 

Ex. 1. Having fully mastered and completed the above, let the 
student take new axes with at the center of the slab. 

Ex. 2. Let him invent a new example with two weights and five 
forces, three of which are unknown. 

54. Case II. Two forces and one co-ordinate are unknown. 

Problem 1. Suppose a circular plate 
weighing 420 lbs., 8 feet in diameter, of uni- 
form thickness and density, to be supported 
in a horizontal position by four wires con- 
nected overhead as in the last illustration. 
See Fig. 47. Two of the forces are known, 
Fi and F 2 ; but two, P and Q, are unknown. 
Fi = 80 lb and acts at A. F 2 = 120 lb. and 
acts at B. One unknown force, P, acts at 
C (x= — 2, y = — 2). The unknown force Q 
acts somewhere on the line HK, whose equation is y = 2. Required 
the magnitudes of P and Q, and the a>cb-ordinate of Q. 
Ans. P = 200 lb, Q = 20 lb, x = -2. 

2. A table is 8 feet square. It has three legs under the points A, B 
and C which are on the perimeter of a 6 ft. square 
as shown in Fig. 48. Two weights are on the table: 
one, W\ is at the center; the other is somewhere 
on the line AD. The leg A supports 80 lb; B sup- 
ports 30 lb.; C supports 48 lb. Find W\ W 2 and 
the position of W?>. 





Fig. 48 



55. Case III. One force and two co-ordi- 
nates an x and a y are unknown. (It is immaterial 
which x and which y are unknown). The data may be tabulated 
as shown on the next page. 

It is assumed that P, x and y are all positive, tho this assumption 
may prove false. In addition to what is given in the Table we know 
that they balance-, hence the three summations are separately zero. 

Find P, x and y, and make a space drawing showing the five forces. 



HOW TO FIND THE RESULTANT 



43 



F 


X 


y 


Fz 


*V 


60 


3 


6 






- 40 


-2 


8 






-104 


4 


+y 






50 


-1 









P 


+x 


-4 




















It is thus seen that the Equations of Equilibrium enable one 
to find the values of three unknowns. A fourth cannot be found 
unless there is some additional condition 
which may lead to a fourth equation, 
such as: T = hP-\-kQ 9 or, x = ay+c. 

It will be seen later, when treating of 
elastic bodies, that some times a fourth 
independent equation can be found. 

56. To find the resultant of a 
system of known parallel forces in 
space, which do not balance. 

It is evident at once that the Equa- 
tions of Equilibrium do not apply. It 
is also evident that the resultant force is the algebraic sum of the 
given forces. It is equally evident that if the positions of the forces 
are given in rectangular co-ordinates, the resultant will at first appear 
as a force acting thru the origin, along the axis OZ, and two moments: 
one about OX and the other about OY. 

That is, we shall have: — 

E = 2F = F 1 +F 2 +F 3 +etc. 
Jfi = S(Fy) =F iyi +F 2 y 2 +etc. 
-M 2 = +2(Fz) = +F 1 x 1 +F 2 x 2 +etc. 

If xi and 2/1 are the co-ordinates of a point in the line of action 
of R in space, we shall have 

-M 2 = Rx 1 = I l (Fx); so that x 1= M^ 



M x = R yi = Z (Fy) ; so that y x = 



R 

VFy 
R 




same point would balance the above system 



This is the general case. It may be 
illustrated by an ideal example. 1?he 
data are most conveniently given in 
tabulated form. 

Show that the resultant is a force 

+2, parallel to OZ, whose line of action 

intersects the plane XY in the point: 

x= 55 

y = 139 

It goes almost without saying, that a 
force of —2 parallel to OZ, thru the 



44 



PARALLEL FORCES IN SPACE 




±Xi 



57. Remarks as to the resultant. 1. If 2(Fy)=Q, there is no 
moment about the axis OX; this shows that the resultant force inter- 
sects that axis. 

2. If both 2(F?/) and 21 (Fx) are zero, the resultant force acts thru 
0, and along the axis of Z. 

3. If 2(F) =0, the resultant is not a single force, but is a moment 
(or a couple of forces).* It will be well to illustrate the last remark. 

58. When the resultant of a system of parallel forces in space is 
a couple. 

1. Ex. The Table represents a set of data, and the calculation. 

Hence M 2 = -Z(Fx) = -10 See (52.) 

M 1 =+X(Fy) = +33 
Since Mi is negative it is laid off on 
the nega- 
tive part of 
the F-axis, 
pointing 
towards 0, 
see 25. In 

this case is a negative angle, or a 
positive obtuse angle. (Fig. 49). 
The resultant couple acts in a plane perpendicular to the line R0, 
with a moment V (1089), which is right- 
handed to an observer looking from R 
towards 0. 

2. A second illustration, in which 
both M i and M 2 are positive. 

M,= 2(/ty) = +18 
if 2 =-2(Fz) = + 6- 
M =V (36 + 324) 
tan 6 = 1/3. See Fig. 50. 

The resultant couple is in the plane AB. 
Exercises. 

Let the student invent and solve two prob- 
lems in one of which M\ and Mi have opposite 
signs; and one in which both M\ and Mi are 
negative. 



F 


X 


y 


Fx 


Fy 


+ 8 


+ 6 


+ 5 


+ 48 

+ 12 


+ 40 


-6 


- 2 


+ 3 


-18 


+ 3 





-8 





-24 


-5 


+ 10 


-7 


-50 


+ 35 


R = 






+ 10 


+ 33 



F 


X 


y 


Fx 


Fy 


15 


+ 3 


+ 2 
+ 10 

+ 12 


+45 


+ 30 


- 6 


+ 8 


-48 


-60 


+ 4 
-13 


-4 
-1 


-16 


+ 48 





+ 13 










- 6 


+ 18 




Fig. 50 



* The formulas give, when R=0, zi=oo and 2/1=00, yet RXxi=oX&=2(Fx) 
which is a finite quantity. Also RXyi=oX<x> —2{Fy) which is another finite 
quantity. The student will see here an illustration of a truth which may have 
seemed obscure in Algebra. 



CHAPTER IV. 

Co-Planar Forces with Center Lines of Action Meeting 

at a Point. 

59. Case I. Three forces balance. Since two non-parallel forces 
acting on a body cannot possibly balance, there must be at least 
three, and any one must directly balance the resultant of the other 
two. This fact gives rise to very important relations between the 
balancing forces which cannot be clearly stated till we have found 
how two intersecting forces may be combined, or replaced by a single 
force. 

1. It is important that the reader picture in his mind the physical 
conditions of the general problem. One body is to be acted upon by 
three other bodies; each action is the resultant of a more or less dis- 
tributed force; these force lines representing concentrated actions 
must lie in a plane; and the lines of action meet at a point. 

2. For example, suppose a heavy solid body, B, rests against the 
combined action of the earth, the smooth end of a block P, and the 
tension in a wire rope Q. The force W is the resultant 
of an attractive force distributed thru a volume. The 
force P is the resultant of a pressure distributed over 
a surface. The force Q is a tension sent along a 
wire from an overhead hook in the plane of W and 
P; the three lines of action meet at 0. The situa- 
tion being ideal, all other actions (atmospheric and 
magnetic) are omitted. P is acting towards 0; W and Q act from 0. 
However, the forces are equally well represented in Fig. 52, as act- 
ing from their common point. 

60. The parallelogram of forces. If two intersecting forces are 
represented by two right lines drawn to scale from or to 
their common point, and if on them as sides, a parallelo- 
gram be drawn, the diagonal drawn from or to the common 
point represents their resultant. A great many proofs of 
this proposition have been given, some of which may be 

' w found in cyclopedias; only one is here given. 

Let the given forces be P and Q, both acting from 0. 
Fig. 53. It is to be shown that the force R represented by the diagonal 
of the parallelogram constructed on P and Q is their resultant. Let C be 

(45) 





THE PARALLELOGRAM OF FORCES 



any point in the plane, and let a perpendicular to the plane at C be 
an "axis of moments." It will now be shown that the moment of R 
about the axis at C is equal to the sum of the moments of P and Q 

about the same axis. Draw OC, and 
connect the arrow heads A, B and D 
with C; finally draw AE and BF paral- 
lel to OC. The force lines being drawn 
to scale, the other lines must be con- 
sidered as drawn to the scale. 

The moment of R with respect to C 
is RX an arm from C to OD. or: 
moment of R = 2X area of A OCD. 




Fig. 63 



Similarly, the moment of P = 2Xarea of A OCA. 

= 2Xareaof A OCE. 



and the moment of 



Q = 2Xareaof A OCB. 
= 2Xareaof A OCF. 



Now, from the equal triangles OAE and BDF it is seen that 
OE = FD, and hence the area of the triangle OCE= the area of the 
triangle FCD. 

Hence the moment of P = 2Xarea of A FCD. 

And since the area OCD= area of OCF-\- area of FCD, it follows 
that the moment of R is equal to the sum of the moments of P and Q. 

But the proof that R is entitled to be called the Resultant is not 
complete, since there are any number of forces acting thru which 
have the same moment. For example, draw thru D a line SS parallel 
to OC. Let D' be any point on >S*S. 

A force 0D' will have the same moment about that R has, since 
the area of the triangle OCD' ' = . the area of the triangle OCD. More- 
over the moment of OD' about any other axis taken on the line OC 
or OC produced, will be equal to the sum of the moments of P and 
Q about the same axis. If, however, we choose an axis outside of 
the line OC, we shall have another line S'S' thru D, and all the forces 
OD' will be rejected except the single force R = OD. 

Hence, since the moment of R is always equal to the sum of the 
moments of P and Q for the axis C, and is the only force which thus 
represents them wheresoever the axis is taken, it alone is entitled 
to the name "Resultant." 

61. The following graphical method of finding the resultant of 
two converging forces, which do not meet within the limits of one's 
drawing, is merely a generalization of the special case with parallel 
forces given in (40). Let P and Q be the co-planar forces which do 



GRAPHICAL CONSTRUCTION OF A RESULTANT 



47 




not meet on the paper. To find the line of action and magnitude of 
their resultant. Fig. 54. 

1. Parallel to the head-and-tail line Ab (which need not be actually 
drawn) draw the head-line BD, and the tail-line aO. 

2. Parallel to the other head-and-tail line, draw 
a head-line AD and a tail-line 60. 

3. The intersection of the head-lines is the head, 
and the intersection of the tail-lines is the tail, of the 
resultant required. 

The analytical proof of this is left to the student. 

62. The resolution of a given force. The given 
forces P and Q are known as the Components of R. 

1. The converse of the proposition in 61 is, that 
any force may be resolved into two components by constructing a 
parallelogram on the line representing the given force as a diagonal. 
Thus Fig. 55, let R be a given force; its com- 
ponents may be Ri and R 2 , but they may as 
well be R\ and iV; in short, there may be any 
number of pairs of components. By drawing 
a variety of parallelograms, the student should 
note the important fact that the resultant of 
two forces may be less than either of them; and 
that one or both the components of a force may 
be greater than the force itself. The relation 
depends upon the size of the angle <f> = AOB. 

2. The sides of the triangle OAD, may 
represent a force and its two components in 
magnitude and direction, but not fully in position; in fact, the posi- 
tion of all the forces might be changed as in Fig. 56, which means 
that the force R is equivalent* to the two com- 
ponents Ri and R2; and conversely, that the forces 
Ri and i? 2 are together equivalent to the single force 
R. As now understood, the lines of this triangle are 
frequently called vectors, a, /3, y and we read the 





Fig. 56 



equivalent equation 



a+/3 = y 



3. Henceforth when we want the resultant of P and Q meeting 
at 0, we shall not draw the full parallelogram, but we shall draw 
0A=P and AD = Q; then OD will equal R. 



* Equivalent so far as the motion or rest of the body acted upon is concerned. 
It is not equivalent when internal stresses are concerned. 



48 



CONVERGING FORCES IN A PLANE 




Fig. 57 



63. The resultant by trigonometry. When P and Q and (j>, 
the angle between P and Q are given, it is evident from Fig. 57 that 

When<£ =tt/2 it is at once evident that 
R = PcosO + Qsind 

and that P sin 6 = Q cos 0, as well as that R 2 = P 2 + Q 2 . 
It is now seen that if the direction and magnitude 

of one component of a given force are given, the 
direction and magnitude of the other component are determined — 
graphically and by trigonometry. 

64. The static triangle or the triangle of equilibrium. We are 

now prepared to bring in the third force which balances two given 
forces. It is evident that in order to balance the two given forces, 
it must balance their resultant. To do this it must have the same 
magnitude and be directly opposite. In other words, referring to 
Fig. 56, and remembering that the third force, which we will call S, 
is the exact opposite of R, and that if R was the vector y, S must be 
the vector — y; so that . n 

or a+/3+y = 

which is the vector equation of equilibrium. The triangle (of mag- 
nitudes and directions only) becomes (Fig. 58) the STATIC 
TRIANGLE or the triangle of equilibrium. 

Whenever three forces balance, such a triangle can be drawn, and 
the relations of sides and angles can be found graphi- 
cally or by trigonometry. The sufficient evidence of 
equilibrium is a state of rest, or of uniform motion. 
Cases of absolutely uniform motion are so rare that 
they are for the present neglected in our thought. 
The great value of the static triangle will be illustrated by a few 
examples. 

65. Imponderable pins. 1. In dealing with centralized forces 
it is often convenient to assume that they act upon an imponderable 
pin, for thus we fix the attention upon a body (the pin) which is at 
rest under their action. This is particularly true when the forces 
arise from the action of wires, ropes, tie-bars, piers, posts, and struts, 
as at the joints of frames. In the case of an over-hanging hoist, such 
as is shown in Fig. 59, we may conceive of a pin of inconsiderable 
weight, acted upon by a tie-bar T (which may be double, consisting 




THE STATIC TRIANGLE 



49 




Tig. 59 




of two thin bars instead of one thick one), a strut, S> or inclined post, 
and a chain or rope which may be connected with a pulley-block 
which has several plies. 

2. The pin stands still; hence the three forces acting upon it balance, 
and we can draw the static triangle, since all directions are given in 
the figure or are known on the construction itself. If the weight, 

W, is known, we can draw the static triangle to scale. 
The values of S and T can be measured from the 
drawing or calculated. See Fig. 60. 

If P represents the pin in Fig. 59, it is well to put 
the same letter within the static triangle, Fig. 60, and 
then follow and note the directions of 
the arrows which surround it. In every 
case the arrow shows the direction of the 
action in Fig. 59. For example: S in 
Fig. 60 points upwards; and it is therefore a strut in 
Fig. 59; T points towards the left; it therefore acts on P towards the 
left, and hence must represent a tie, or tension bar. 

3. If three non-parallel co-planar forces act upon a body at different 
points and yet balance, the three lines of action must meet at a point, 
or one could not directly balance the resultant of the other two. 

66. Ex. 1. Take the ideal case of a heavy bar suspended against 
a vertical smooth wall. A weightless wire fastened to a hook or staple 
in the face of the wall at C supports the end 
By Fig. 61, while the other end rests against 
the smooth wall which can act only in the direc- 
tion of a normal to its surface. The bar and 
wire must be in a vertical plane normal to the 
wall. If G is the center of gravity action of 
the bar, the line of the wire must pass thru the 
intersection of the action lines of H and W at 0. 
The static triangle can now be drawn showing the magnitudes of forces 
and angles. Fig. 61a. 

2. If G be at the center of the line AB (as in the case of a uniform 
bar) it is seen that is the middle point of CB, and AD = AC. Hence, 
when 6 and /, the length of the uniform bar, are known, the position 
of C is found by laying off AC equal to I cos 9. 

3. The reader must not fail to see that the above suspended body 
is in unstable equilibrium; the touch of a fourth force would cause 
it to fall or to collapse against the wall, even if it were prevented from 
swinging sidewise. If, however, the wall were rough (and all walls 
are rough), there would be a certain stability. 




50 CONVERGING FORCES IN A PLANE 

4. It is evident that the pull of the wire CB could be replaced 
by the thrust of a strut or prop SB, without in any way 
changing the static triangle; we would then let T stand for 
thrust instead of tension. It is also evident that a heavy weight 
may be hung at G without dislocating the bar, as it would only 
increase W. 

G7. A third example will further illustrate the principle that 

three balancing forces meet at a 

/' X point. A slender uniform rod rests 

/ /cHa\ ^ 3 upon the inner surface and edge 

/ . on J i a \\ ^^ of a smooth hemispherical bowl. 

, — I ../. -^--..-XJL|p() pig. (j2 # We are to find the relation 

\ A \^*\ /\ between I and 0. 

\^Hq i !/ ^ ne ac ti° n at A must be normal 

^/^. to the spherical surface, and hence 

along a diameter; the action at D 
must be normal to the rod, as tho the edge were an infinitesimal round. 
Hence the gravity action must be in a line vertically below 0, where 
the two normals intersect, and the length of the bar, which will rest at 
A and Z), is determined, for its center must be at G. 

1. If be the inclination of the rod to the horizontal, it is evident 

from the figure that 2rcos2# = - I cos 6, which gives the necessary 

relation between I and 0. It is further evident that any movement of 
the rod due to a new force would have the effect of raising the point 
G, or lifting the bar, so that when relieved from the disturbing force 
it would slide back to its position of equilibrium. This is a case of 
"stable equilibrium." 

2. The inferior limit to the length of the rod we have been dis- 
cussing is when l = %rcos0. The minimum length of the rod is found 
by eliminating from the two equations 

2rcos2 = — IcosO 

/= £rcos# 
Whence Z = 2r V(2/3) and cos#= 1W6- 

3. If I be less than |rV6. the bar will rest only when G is directly 
below C. 

Problems. 

68. 1. Given a smooth vertical wall, and at a distance s a 
lower parallel wall with a horizontal edge which is rounded and smooth. 



GRAPHICAL SOLUTIONS 



51 



Fig. 63. What is the inclination of a smooth, slender, uniform rod 
which will rest over the corner and against the wall? 
(Note where the balancing forces meet). 
Ans. 



cos 



6= «?*. 



2. 





Fig. 63 



Find the stress in the members of this derrick due 

to the weight (8 tons). 
Fig. 64. 

Draw a static triangle 
for the pin at C, and then another for 
the pin at B. 

3. If AB and AB' are equal smooth 
rods looped on a pin or ring at A, 



Fig. 64 



Fig. 65, prove that the condition, that the pair 
will rest on the surface of a smooth horizontal 
cylinder whose radius is r, is 



1 = 



sin 2 tan 6 



4. 



A smooth bar rests upon two smooth in- 
clined planes 





Fig. 65 



whose intersec- 
tion at is hori- 
zontal. Find the inclination of the bar 
to the horizontal. Fig. 66. 

The weight is centered at G. First 
draw the static triangle of external 
forces W, Fi and F 2 ; Fig. 66, b; then 
resolve the weight into vertical com- 
ponents, W\ at P, and W* at Q, and 

complete the static triangles for the points P and Q, thereby finding 

CD, the "thrust" in the bar, and its inclination. 

From the diagram (b) prove that 



tan 



n_ a cot a — b cot ft 



5. A heavy cylinder is held against a smooth wall by a smooth 
rigid bar which is hinged upon a pin at A (Fig. 67), and held by a 
tie T to a hook at H, as shown. Given the weight and radius of 
cylinder, the length of bar, and all angles, to find the tension T, the 
pressures Pi and P 2 , and the magnitude and direction of the hinge 
action, due solely to the weight of the cylinder. 



52 



CONVERGING FORCES IN A PLANE 




Solution. Three forces act upon the cylinder, hence the static 
triangle KEF whereby Pi and P 2 are known. Fig. 67. 

Since we omit the weight of the bar A B, only three forces are acting 

upon it, viz.: Pi, T and the hinge. 
We already have Pi, and we know 
the direction of T, and as the three 
must meet, say at D, we know the 
direction of the hinge action, P 3 , 
namely in the direction AD. Its 
static triangle is therefore FEH. 
Thus the magnitudeof Tand both the 
magnitude and direction of the hinge 
action due to the cylinder are found. 

09. How to determine the tension or compressive force in a bar 
of a loaded frame from the static triangles of its pins. 

Let the members of a cantilever frame lie in a vertical plane with a 
load at the end. Fig. 68. We are to find 
the actions of the bars BA and CA upon the 
pin at A; and the actions of the bars AB, CB 
and DB upon the pin at B. 

Solution. The static triangle for A. The 
case is similar to that in 59. Knowing one 
side of the triangle and the directions of the 
other two, it is quickly drawn. W acts down, 
BA acts up, parallel to the bar, and CA acts 
up to the left parallel to the bar CA, closing 
the triangle. The static triangle for A shows 
magnitudes and character. AB acts up on 
the pin A, and is therefore, a strut ; the bar 
CA also acts up and away from the pin, and therefore is a tie. 

Going now to the pin B, three bars act on it and balance; hence 
B has a static triangle. The action of AB is known, being a strut it 
acts down on B, and the triangle already drawn gives it magnitude. 
We therefore use the line ba, or ab, and draw db for the bar DB, and 
cb for the bar CB, and mark the arrows of the static triangle for B. 
We note that the arrow db points to the right, showing that DB acts 
against the pin B, while the arrow cb points up, showing that CB 
acts away from the pin B. DB is therefore a strut, and CB is a tie, 
and they are lettered accordingly. 

This is all very simple, but it cannot be too clearly seen. Occasions 
will surely arise when students will find it necessary to refer back to 
this section and go over it again. 




Fig. 68 



THE STATIC POLYGON 



53 




TO. Case II. The polygon of forces and the static polygon. 

When more than three co-planar forces are given with a common 
point in their lines of action, their resultant, or their necessary re- 
lations, if they form a balanced system, are readily found 
by an extension of what has already been given. Fig. 69. 
Given five forces acting upon a pin at P. To find 
their resultant. 

Graphical Solution. 

Fig. 69 

Take any convenient point as an origin; draw OA 
equal and parallel to Fr, draw AB equal and parallel to F 2 ; this gives 
OB equal and parallel to the resultant of F\ and F 2 . 

Draw BC equal and parallel to F 3 ; OC is then 
equal and parallel to the resultant of Fi, F 2 and F 3 . 
Draw CD equal and parallel to F 4 ; OD is then 
equal and parallel to the resultant of Fi. . . . F 4 . 
Draw DE equal and parallel to F 5 ; OE is then 
equal and parallel to the resultant of all the forces, 
F, . . . . F 5 . 

The order in which the forces are taken is not im- 
portant, the resultant is still the same. This state- 
ment should be put to the test by a careful drawing. 
It is evident that the lines 0B 9 OC, etc., are of no 
use excepting always the last one which is the resultant 
sought. The line representing their resultant is always drawn from 
to the end of the last component force. 

Had the line of the last component ended in 0, there would have 
been no resultant, as we should have had 

R = 

which is the evidence of a balanced system of forces. In all practical 
statical problems R is zero; and every polygon drawn as above, which 
closes, is called a Polygon or Equilibrium, or more conveniently a 
"Static Polygon." 

71. Static triangles and polygons are useful in finding unknown 
magnitudes and directions. In every case where a resultant is to be 
found, or a static triangle or polygon is to be drawn, the data must be 
geometrically sufficient to that end. In most of the problems thus 
far solved, the magnitudes of two forces were found, all angles being 
given or determined by the geometrical necessity of convergence 
of the lines of action. In the last general problem several forces 
were fully given, and a single angle and force were to be found. The 
following problem is in a new class. 




Fig 



54 



CONVERGING FORCES IN A PLANE 




Fig. 70 



Ex. Given one force in position, a second force in magnitude 
only, and a third in direction only. 

This may be illustrated by a special device using weights (Fig. 70) 

F u F 2 and F 3 . 

Let be the pin or 
body acted upon. 

Let Fi represent the 
forcefully known =4| 
acting down. 

Let a be the known 
direction of F 2 > relative 
toFi;a=150°. 

Let F 3 = 3 represent 
the magnitude of the 
third or balancing force. 
We lay off to scale 
from any point as A, 
Fi as AB to the point B; draw the direction of F 2 as 0D; with F 3 as a 
radius, from A as a center, draw the arc C'C". The intersections of 
this arc with the line BD determine the magnitudes of F 2 ' and F 2 ". 
We get in general two solutions for the magnitude of F 2 and two 
directions for F 3 . We see that F u F 2 and F 3 ' balance at 0; 70a, and 
that F u F 2 ," F 8 " balance at 0'; (6). 

72. Illustrations of the use of the static polygon. 
Ex. 1. Four equal cylinders lie within a hollow semi-cylinder as 
shown in vertical end view. Fig. 71. Given 
the weights, to find the pressures Pi, P 2 , 
P 3 , P 4 , P 5 , Pe, and P 7 . 

Beginning with A, we draw the static tri- 
angle WiPiP 2 . Fig. 72. We next draw the static 
polygon for B: W 2 P 2 P 3 P^ Then follows 
the polygon for C: W 3 P,P b P 6 . 

Finally, the static triangle for D : W±P%Pi, and the polygon 

w k closes. The reader will notice that since the four forces acting 

on B balance, they must, when drawn as vectors, form a 

w 3 closed 'polygon; hence P4 returns to 0. Having drawn 

Fig. 72 accurately, we should have no difficulty in check- 

: ing the following: 

P 7 = P 1 = W^2- V2 

P & = P 2 = W 

P 5 = P3 = JF(1+V2)V2- V2 
p i= W^% 




Fig. 71 




EQUATIONS OF EQUILIBRIUM 



55 



Ex. 2. A heavy horizontal cylinder, solid or hollow, rests 
two unequal cylinders which rest upon a horizontal bed with 
walls. Fig. 73. Weights being 
given, find the pressures at all 
points. The horizontal thrusts 
in AB and AC being balanced 
by retaining walls. 

The diagram (Fig. 74) 
is exceedingly simple, 
and values may be cal- 
culated when weights 
and radii are known. 

Of course it was evident without the drawing that 
Pt+P^Wi + Wz + Ws, 



upon 
guard 



w a 





/— ^m 


7777777/7/// 
/ 


I W 3 




p 4 




— tL__b\ 


w 2 \ 


f***** 


j 


J 


i P 3 
| 


\\\m 


1 Fig. 73 


*4 



and that P 3 
the group. 



P 6 , as they were the external forces acting on 



73. The static polygon in "graphical statics." The Static 
Polygon enables us to find the magnitude of two forces when 
all other "parts" are known. This property is of the greatest 
value in finding the stresses in the members of frames when 
both frames and external forces may be regarded as co-planar. This 
deserves a somewhat extended illustration, as it will bring to us 
the very elegant method of "Reciprocal Polygons"* given in Chap- 
ter XI. 

74. Equations of equilibrium and their uses. In every problem 
thus far solved by means of the static triangle or the static polygon, 
the data (forces and angles) were assumed as known, and the accuracy 
of the drawing was readily checked by trigonometry applied to the 
diagrams. When the known angles are the inclinations of forces to 
a given direction, that direction should be taken as the axis OX with 
OY perpendicular to it, and the rectangular components found by 
the method already given. Since the forces balance, their com- 
ponents along OX and along OY must balance independently, and 
their moment about any axis must be zero. Hence the Equations 

of Equilibrium ^^ a 

luh cosu = 

XFsinO = 

2F/ = 

Here F is the representative of a force, known or unknown, and 6 is 



* So named by the eminent scholar Clerk — Maxwell. 



56 



CONVERGING FORCES IN A PLANE 



its inclination to the axis X. For brevity we may write Fi = Fcos0 y 
and F 2 = F sin 0, so that the equations become 

HF l = 
2F 2 = 

HFl = 

The angle 6 is measured positively from -\-X, clockwise, to the head 
of the force arrow drawn from 0. 

The arc 6 is always positive and its valve lies between 0° and 360°. 

75. Equations for a resultant. When the given forces do not 
balance, the resultant is readily found by means of the following 
equations, which, after the above, are almost self-evident. 

HF^R, 

ZiF 2 = R% 




R = ViJi' + JRa 2 

tan = R 2 /Ri 

All these quations are useful, especially those of 74. The third 

equation 2FZ = 0, is convenient when I is the perpendicular from a 

point on the line of one force to the line of action of another force. 

Ex. 1. A rigid prismatic rod of small diameter stands on a rough 

horizontal plane leaning against a smooth vertical wall. 

Find the pressure against the wall. Fig. 75. Three 

forces are acting on the rod, and they balance. Hence 

their lines of action must meet at a point. Given the 

position and length of the rod, the point of convergence 

is seen to be where H and W meet. If we take B 

Fig. 75 as the axis for moments we see at once, since HFl = 0, 

#Zcos0 = iJ^Zsinf9 
H = \W tan 

But this value can be found from the equations SFi = and 2F 2 = 0. 

H-Ps'm/3 = 
W-Pcosfi = 
Hence #/FF = tan/3 

But from the figure tan/3 = J tan#, hence 

H = JPFtan0, as before. 

This shows that the three equations of 74 are not independent. They 
serve for the discovery of but two unknown quantities. 



A CLUMSY SOLUTION 



57 



F 
16 
20 
10 
25 





cosO 


s'mO 


F 1 


F 2 


7T/4 

140° 


















210° 










3. 77 
2 











R = 



e,= 



Ex. 2. Find the resultant of the forces given in the table, both 
by graphics and by analytics; the forces all act at a point in OX, 

76. Comparison of graphical 
and analytic methods. All of the 

problems we have had thus far 
in this chapter may be solved by 
equations of equilibrium. We 
shall have frequent use of these 
equations later on, and the stu- 
dent should finally be equally free 
to use the graphical method or 
what is called the analytical 

method. In earlier works on Mechanics, graphical methods were little 
used. For the sake of comparing the method by equations alone with 
the graphical method with trigonometry applied to the force polygon, 
we quote from good authorities solutions of two prob- 
lems already solved. 

Ex. 1. "A heavy uniform beam, Fig. 76, AB, rests 
with one end, A, against a smooth vertical wall, and 
the other end, B> is fastened by a string, BC, of given 
length, to a point, C, in the wall. The beam and the 
string are in a vertical plane; it is required to deter- 
mine the pressure against the wall, the tension of the 
string, and the position of the beam and the string. 

"Let AG = GB = a, AC = x, BC = b, 




Fig. 76 



weight of beam = W, tension of string = T, pressure of wall = R. 




BAE = 0, BCA=(f> 
then we have 




for horizontal forces, R — T sin <f> = 


(i) 


for vertical forces, W—T cos <j> = 


(2) 


for moments about A, Wa sin 6 — ( T.AD = Tx sin </>) = 


(3) 


.*. a sin 6 = x tan <\> 


(4) 


and by the geometry of the figure 




b sin# 
2a sin<£ 


(5) 


x sin (0 — <f)) 


(6) 



2a 



sin 



4> 



58 



CONVERGING FORCES IN A PLANE 



Solving (4), (5) and (6), we get 



x = 



COS(f) = 



// 


-4a 2 


i 

2 


- 


3 




2 


fc 2 -4a 2 


h 


3 





SI 



\\\9= — 



l 

2a 



16a 2 -b 2 
3 



from which R and T become known." (Price's Anal. Mech's, Vol. I 
p. 69.) 

Ex. 2. For the second example, showing the occasional simplicity 
of the analytic method, I quote a problem and solution from Pro- 
fessor Edward A. Bowser, Analytic Mechanics, 1884. 

"A heavy beam, AB, rests on two given smooth planes which are 
inclined at angles, a and ft, to the horizon; 
required the angle which the beam makes 
with the horizontal plane, and the pressures 
on the planes. 

"Let a and b be the segments, A G and BG, 
of the beam, made by its center of gravity, 
G; let R and R f be the pressures on the planes, 
AC and BC, the lines of action of which are 
perpendicular to the planes since they are 
smooth, and let W be the weight of the beam. Then we have 

for horizontal forces, R sin a = R' sin ft 

for vertical forces, Rcos a-\- R f cos f3= W 

for moments about G, Ra cos (a — 0) =R'b cos (fi+0) 

Dividing (3) by (1), we have 

a cot a + a tan 6 = b cot ft — b tan 6 




rig. 77 



(i) 

(2) 
(3) 



n acota — bcotB 
tanc7= '— 



a + 6 



and from (1) and (2) we have 

Ws'mft m 
sin(a+/3) 



R = 



R' = 



Wsina 



sin 



("+£) 



77. The elegance and simplicity of the force diagram. Two per- 
forated balls, A and B, whose weights are W\ and W 2 » slide down on 
two straight smooth rods until further motion is prevented by a weight- 



FORCES CONCENTRATED UPON PINS 



59 



less thread between them. The rods are stretched in a vertical plane 
from the same pin with equal inclinations a. Prove that 
Wx-W* 



tan# 



cot a. (See Fig. 78), 



and that the tension in the thread is 

W!-W 2 



T = 



2 sin 9 




Fig. 78 



Problems. 



2. 



1. A triangle frame with pin-joints, whose sides are 5', 6', 7', stands 
vertically with its shortest side resting on a smooth level floor. What 
stress does a weight of 269 lbs. placed at the vertex cause in the lowest 
side? Draw the static diagram and measure the force line. 

A camp stool (Fig. 79), consisting of four equal legs arranged 
in two oblique crosses, as AK and CH (with a hori- 
zontal connecting pin at 0, and parallel bars con- 
necting corresponding tops of the legs at A and C) 
carries a flexible band ABC from bar to bar as 
shown. The band supports a heavy rod of small 
diameter, B. Assuming that the stool stands on a 
smooth horizontal floor; that all the members of the 
stool are without weight, and that there is no fric- 
tion at joints, we are to find the position of equili- 
brium, when the weight of the rod, and the lengths 
of legs and band are given. 




Pig. 79 



78. The concentration of loads upon the pins of a frame. 

Rule. When a member 
of a frame has weight, and 
whenever it carries con- 
centrated loads, both the 
weight of the member and 
the loads on it must be 
resolved into parallel 
components acting on the 
two supporting pins. 

The derrick shown in 
Fig. 80 has a cantilever or 
overhanging beam with a 
concentrated load, W 2 , at the 
outer end. The weight of 
the beam, W\, and W 2 are to be resolved into components at the two 




Fig. 80 



CONVERGING FORCES IN A PLANE 



pins, F and E. Then the static diagram can be drawn, 
center of gravity of the beam; let W\ be its weight, W\ 
ponent at E; and W\ the component at F. 



G is the 

its com- 



W l , = — , W l9 

a + b 

In like manner W 2 is resolved into 



and Wi"=-^- W x 

a + b 



and W 2 "= + 



a + b 
a + b+c 
a + b 



W 2 at E 



W 2 at F. 



The transformed derrick now is as follows: The load at F is W 2 " + 
W\ (Fig. 80a), and the static triangle for pin F is 
ABC (Fig. 186), which gives the thrust P in the boom, 
and the tension in the beam T. The forces now known 
as acting upon the pin E are the horizontal tension 
T and the negative load W 2 — W\\ the resultant of 
these two forces is shown in Fig. 80c to be the force 
DH acting upwards. The static triangle for pin E 
is now readily drawn, viz., DH up (already found); 
HK down, parallel to the stay S; and KD up, parallel 
to the mast N; giving for E the triangle of forces 
DHK; and all the forces or stresses are found. 

If instead of the resultant DH we had used its components T and 
(W2—W1), see Fig. 80c, we should have had the Static 
polygon of four sides, DJHK. We shall soon have static 
polygons in abundance. If all angles are 
given, the magnitude of the forces, in 
stay, mast and boom can be calculated 
by trigonometry. 




Fig. 80(a) 





Fig. 80(b) 



79. Stability due to friction. 1. When 
a light bar, like a cane, a ladder, prop, 
or strut of any kind, transmits a thrust against a rough plane surface, 
the roughness of the plane tends to hold the end 
of the bar against slipping. Fig. 81. If the 
thrust is normal to the plane, there is no tend- 
ency to slip and the roughness is not needed to 
keep the bar in position. When, however, the 
thrust is inclined to the normal like P, Fig. 81, 
the resistance R must also be inclined, so that if 
it also be resolved into rectangular components, 
one of which is normal and one tangential, along 



Fig. 81 




STABILITY DEPENDENT UPON FRICTION 61 

the plane, the latter is Psin#, and the normal component is PcosO. 
The ratio of the tangential to the normal component is 

Psin9 , /) 
2j =tanc7 

%. If the inclination (to the normal) be gradually increased, a value 
of 9 will soon be reached beyond which the roughness of the plane will 
not suffice to balance the tangential component of P, so that if 6 
is made any larger, the bar will slip and fall. The static angle is 
therefore limited, and the limiting value of 9 is called <j). The force 
thus exerted by a rough plane in the direction of the plane itself is 
called Friction, and <£ * s called the "Angle of Repose" or the "Angle 
of Friction," or perhaps the "Static Angle." The roughness of sur- 
faces both of bars and planes varies greatly. 

3. Another way to illustrate friction is to place a rough body 
upon a rough inclined plane, Fig. 82. If it stands 
still, the forces balance. The normal component of 
the weight is balanced by the normal action of the 
plane; Wcos6. The tangential component of W is 
JFsin 9, and this must be balanced by the friction.* If 
9 is small, very little friction is utilized or developed; 
as 9 is increased more friction is needed. When 9 = (b, 

Fig. 

the limiting value of friction is reached for that body, 
and the body is upon the point of sliding. If # > <£ it will slide if 
a little initial sticking is overcome by a slight disturbing force. At 
the "static angle" 

Friction — PFsin (j) = tan <f> (Wcos <f>) =N tan <fc =fN 

When#></> the component of W down the plane becomes greater, 
while the total friction usually becomes less, since N = Wcos 9 is less. 
Hence there is no longer a static body. 

4. The factor tan<£, standing with or before A 7 , is called /, the 
"Co-efficient of Friction," which is used when the body is actually 
moving. When a body is at rest and in no immediate danger of 
moving, the frictional action is less than Ntancf), and it may be zero. 
It is, however, of the highest importance that the direction of the 
oblique force which is to be balanced by friction be known and pro- 
vided for. The following problem will illustrate this point. 




* The reader will observe that the component of W parallel to the plane acts 
thru G, and forms with the friction, fN, a couple which when combined with 
Wcos#= N brings the center of pressure to the point C. 



62 



CONVERGING FORCES IN A PLANE 



Problem. 

80. A ladder, weighing 60 lbs, stands on a rough level floor, 
leaning against a smooth wall at a point 24 feet above the floor, with 
its foot 12 feet from the wall. Fig. 83. If <£ = arc tan 1/3 be the "angle 
of friction" between the ladder's feet and the floor, how far up the 
ladder can a man weighing 180 lbs. go before the ladder will slip 
and fall? 

1. Before the man steps on the ladder, the line of action against 
the ground is not straight down the ladder; it is nearer the normal, as 

seen in Fig. a, where 
tan = 6/24. The three 
forces acting on the 
unloaded ladder are: a 
horizontal force AH; a 
vertical force OC = 60 
meeting AH at C; the 
third force, CB, namely 
the action of the floor, 
must pass thru C; its 
obliquity is 0. 

2. When the man steps 
on the ladder, the point C is over the center of gravity of the two 
weights, the ladder and the man. Hence, it follows that, at first C 
has moved out, and the ladder is less likely to slip. So long as the 
man is on the lower half of the ladder, it is stable. Thus, when the 
man stands on a rung six feet from the floor, at S (Fig. b) the center 
of gravity of the two weights is 8.25 feet from the wall, and 




tan# 



3.75 
24 



3. When the man is at the middle, tan # = 6/24, as when the ladder 
was unloadedc 

As the man mounts higher, the center of gravity moves to the left 
as does the point C, where the three forces always meet. 

When the man reaches the point £ (Fig. c), three-fourths of the way 
to the top (supposing that possible), the center of gravity of the two 
weights (and therefore the point C) has moved to a point only 3.75 



feet from the wall, and tan# = 



8.25 

~24" 



which is more thantsai(j>. Hence 



the ladder slips and falls just when the man is about to reach a height 
of 18 feet. 



WHY THE LADDER FALLS 



63 



4. Experienced workmen know well that ladders are apt to slip 
if a man (and the heavier he is the greater the danger) goes too near 
the top, and by sad experience they have learned to secure the foot 
of the ladder, either by a block or tie, or by putting another man on 
the ladder near the foot so as to bring the center of gravity of the 
whole weight away from the wall thereby diminishing the angle 6. 

5. Workmen know, too, that a light ladder is more likely to slip 
than a heavy one. If we were to make the unreasonable assump- 
tion that the ladder has no weight worth considering, the point C in 
the line of action of the supporting reaction would always be directly 
over the head of the man as he mounts the ladder, and in a horizontal 
line from the top of the ladder. 

6. If the student will draw the static triangle for each of the three 
positions of the man on the ladder, he will see how H increases as the 
man mounts. 

81. While friction is a great hindrance to the motion of bodies, 
it is a great aid to the stability of bodies at rest, 

1. It was assumed above that the wall was absolutely smooth. 
Hence it could offer no resistance to the slipping down of the top of 
the ladder. In reality, as all walls are rough, a tendency to slide 
down would have developed an upward force equal to Z/tanc^, (<^2 
being the "Angle of Friction" for the ladder and wall), which would 
have had the effect of moving the center of gravity of the weights 
away from the wall thereby diminishing 6 (and perhaps saving a fall). 

The student should clearly see how much H tan<£ 2 = #/6 would 
move C away from the wall pro- 
vided the resultant load was act- 
ing at a distance of 3.75 feet from 
the wall. This matter is worth 
a bit of careful study. In fact, 
the value of H depends some- 
what upon the friction between 
the ladder and the wall. 

2. The center of the ladder 
(Fig. 84) is at G with a man at S. 
The resultant load is at R, 3.75 
feet from the wall. When the 
wall is smooth the point C is 3.75 

feet from the wall and the reaction of the floor is along the line BC 
so that 

tan<9= — =0.344 




Fig. 84 



64 THE STABILITY DUE TO FRICTION 

Figure (84) (a) is the static triangle for the ladder, the value of H 
being 

240 X ^- =82.5 lbs. 

24 

If, now, we suppose the wall is rough, we shall have the co-efficient 
of friction / = tan<^>2. If we suppose the ladder is upon the point of 
slipping, the frictional action of the wall, which is up, is fH\ so that 
we now have a new force involving H', acting upon the ladder. To 
find the value of H f under new conditions, take moments about B 

24#' + 12j77'-6X60 -9X180 = 
from which we get inoft 

H '= J™L = 76A lbs. 
if/ is 1/6. 24 + 12 / 

This gives the value of H' less than former H, so that the static triangle 
is new, the base being the new value H r , and the vertical line being 
24>0—fH' the new 0' is seen to be less than the former value of 6. 



_2_ 

tan0' = -^?- -0.335 



76i3 



■/A 



The movement of the point C to C", due to the friction on the wall, 
is readily seen if we get the resultant of the two vertical forces, namely 
240 lbs at R, acting down, and H' acting up at the point A. By what 
was given in a previous chapter, the resultant is the force acting at 
the point below R, with a magnitude of 240 lbs. —fH', giving the new 
point C graphically. In fact, it is enough to prove the point stated 
above, namely, that the point C is carried further away from the wall 
by the roughness of the wall, since the resultant is sure to be beyond or 
below the point R. We thus see that the stability 
of a body, like the leaning ladder, may be secured 
by means of friction at its base, assisted by friction 
at its top. 

82 • Again, to use a modified form of the prob- 
lem already solved twice ( ) of a uniform bar 
suspended against a vertical wall. It was found 
that when the wall was smooth, the cord or wire 
should be made fast to a hook as much above H 
as H was above D. Fig. 85. If now the wall be 
rough with an "angle of friction'* <f>, the bar (and 
its load hung at G) will be in equilibrium if C (where the three force 
lines meet) be anywhere between E and F; tho the safest place is at the 
center, C. 




CONCENTRATION AND RESOLUTION 



65 



\ 

1 


\ 

) 




1 



Fig. 86 



83. The concentration of distributed forces and the resolution 
of concentrated forces. When a heavy homogeneous and uniform 
timber has a support directly under or over its geometrical center, 
Fig. 86, we say it is in equilibrium, because its 
center of gravity is supported. We may say 
more explicitly that the resultant of the earth's 
attraction, which is distributed thru the whole 
mass of the beam, acts thru the center, and that 
the resultant is balanced by the support. But 
one may ask, how does the attraction on the ver- 
tical end layer get to the center? We answer that 

the first layer hangs by cohesion on the second layer; and that upon 
the third; and so on, each layer holding up all between it and the 
end, and being held up with its load by the one next to it on the inside. 
This vertically-acting force, made possible by cohesion, has been called 
"shearing stress." Like friction, it is tangential, but unlike friction, 
it is far more intimate and continuous, the molecular interlocking being 
nearly perfect instead of imperfect and haphazard. By means of this 
shearing stress, the weights of all parts of the timber are brought to the 
center. The shearing stress is evidently the greatest at the center. 
The fact that this transfer of weight develops other stresses within 
the timber does not now concern us, tho it will concern us later on. 

8 4. If the timber has two supports, Fig. 87, the weight is con- 
centrated at two points. The simplest method of making this dis- 
tribution is by first concentrating the whole 
at the center of gravity, and then resolv- 
ing W into two parallel components at A 
and B, in accord with the principles of 
Chapter III. 

Since the moments of the components are together, equal to the 
moment of their resultant about any axis, take an axis at A to find 
the component at F 2 ; then 

Wa = W 2 (a + b) 

hence V 2 , which must directly balance W 2 , is found to be (in mag- 
nitude) 



w* 



w 

Fig. 87 



V 2 = 



a + b 



W 



In like manner 



Fi = 



a + b 



W 



85. Sometimes the method of support brings into a member of 
a frame a direct stress which is independent of the shear and bend- 



66 



CONVERGING FORCES IN A PLANE 



ing stresses, as in this simple case. Suppose this heavy timber is 
supported by two separate cords or chains which are connected to a 
common hook. Consider the chains and timber as forming a tri- 
angular frame, with the weight of the timber 
(and whatever it may carry) concentrated 
at the ends into W\ and W 2 with a single 
support at the hook. The static 
diagram can be drawn in the now 
familiar way, so arranged that 
only four straight lines exhibit 
the static triangles for the pins 
A and B. The cords are sub- 
ject to direct tension, the timber 
is subject to a direct thrust, and to shear. Fig. 88(a). 

Had the timber and its load been supported by two chains 
to separate hooks, the timber might have been subject to direct tension. 
The student may so represent a timber and draw its static (or 
stress) diagram. 





CHAPTER V. 

Non-Concurrent, Co-planar Forces. 

8G. In order to determine fully the force or couple required 
to balance a system of co-planar forces which act in different direc- 
tions and at different points of a rigid body or frame, the resultant 
of the given forces is first found. There are four rather distinct 
methods of doing this. Every force must be actually or ideally 
given in magnitude, direction and position. 

I. Graphical solution. Fig. 89. Combine the forces in pairs 

by producing their lines of action, if need 
be, till they intersect, and using the paral- 
lelogram, or by the methods of IV., £ 
and (3). Combine the partial resultants 
in pairs, and so on till the single resultant 
is found. A little forethought in forming 
pairs will generally keep the work within 
bounds. The student should take his 
data from what is given in Fig. 89, as to 
the forces (1), (2), (3) and (4), their 

directions and their relative positions. The axes OX and OY are not 

needed in this method. 




Fig. 89 



THE ALGEBRAIC SOLUTION 



67 



If the final pair should consist of two equal and directly opposite 
forces, the resultant is zero, and the given forces balance. 

If, in any such problem, the last pair should chance to form a couple, 
the combination can go no further; the resultant is that couple, and 
the system cannot be balanced by any single force; only an opposite 
couple can prevent the body from turning. 

87. II. Analytic solution. Choose a pair of co-ordinate axes 
which can conveniently be used in locating points of action and direc- 
tions of forces. Let the direction 
be determined by the angle 9 as 
shown in Fig. 90, and let a point 
of application be given by x and y. 

(a) Reslove every force into 
its components Pcos$ = Pi, and 
Psin(9 = P 2 . See Fig. 90. 

(6) Resolve every component 
into an equal and parallel force 
at 0, and a couple whose moment 

(about OZ) is -{-xP 2 for one component, and —yPi for the other. 
(Fig. 90.) Thus the resultant moment for a force P is xP 2 — yP 1 . In 
the same way every given force, in general, is resolved into two forces, 
Pi and P 2 , at 0, and a moment about OZ. 

(c) Summing results 

iJ 1 = SPi = SP cos 0. 
P 2 = 2P 2 = 2P sin 0. 

M = X(xP 2 -yPi) =Sa?P 2 -SyPi. 




Fig. 90 



(d) Combining Pi and P 2 gives R in magnitude and direction, but 
not in position. 

R= VP! 2 +P 2 2 , 

Pi 



tan^ v = — 



(e) Combine M and R by 33, and find the position of the resultant 
force, using the equation p = M/R. The balancing force is of course 
the above resultant reversed. 

(/) The point of application cannot be explicitly found, but the 
line of action can, and any point in that line can be taken as the point 
of application provided it be a point in or on the body upon which 
the given forces act. 

If x' and y' are the co-ordinates of a point in the action line of P, 
we have the equation of that line 



M = x'R 2 -y'R 1 



68 



NON-CONVERGENT CO-PLANAR FORCES 



whose intercepts on the axes are readily found and the action line can 

be drawn and the value of p be checked. 

This method can be made clear by two examples. 

Ex. 1. Find the resultant of the forces shown in 
the following table which gives magnitudes, co-ordi- 
nates of points of application, and directions of forces. 
The table also shows how to tabulate the work of cal- 
culation. Only two decimals are used in the natural 
sines and cosines. The forces shown in Fig. 91 are 

fairly represented in the table. 




F 


X 


y 

-4 


e 

45° 


F 1 


F 2 


xF 2 


yFi 


10 
12 

8 


-5 


+ 7.07 


+ 7.07 


-35.35 


-28.28 



2.5 


-6 


120° 


- 6.00 


+ 10.39 


0.00 


+ 36.00 


-1 

2 


300° 


+ 4.00 


- 6.93 


-17.33 


- 4.00 


16 


2 


180° 


-16.00 


0.00 


0.00 


-32.00 


6 


4 


1 


40° 


+ 4.62 


+ 3.84 


15.36 


+ 4.62 




/?, = -6.31 


#2 = 14.37 


-37.32 


-23.66 



Results: see Fig. 91 

R=(R 1 2 + R 2 2 ) l = 15.63 



9 = arc tan 



Ri 



ii3°40'. 



V = M/R= -0.87. 



M= -37.32 + 23.66= -13.66. 

Ex. 2. Since both R\ and R 2 are zero, 
the resultant is not a single force, but a 
couple whose moment may be found. 

88. III. Semi-graphic method. This 
requires an accurate drawing of the force 
lines so that in each case its perpendicular 
distance from some convenient central 
point can be measured. If possible, 
should be at the intersection of two force lines. Call the perpendi- 
culars p± 9 p 2 , etc. 

(a) Imagine that every force F is resolved into an equal parallel 
force at 0, and a couple whose moment is Fp. Measure the perpen- 
diculars and calculate the value of M = HFp. 

(b) Beginning at (or some more convenient point), draw the 
force polygon of the transferred forces which meet at 0, thereby 
finding the magnitude and direction of the resultant R. 



F 


X 


y 


9 


24 


3 


-l 


90° 


14 


-8 


+ 10 


300° 


40 


4 


4 


135° 


45.5 


+ 6 


-6 


298°2 / 



THE CHAIN POLYGON 



69 



(c) Combine the resultant force R with the resultant moment M, 
as already pointed out in the last method, and find p. 

The student should use again the data shown in Fig. 89, and 
in the Table under the II Method, and so check up all former 
results. 

If in any new problem if = 0, and i?>or<0, the final resultant is a 
force R acting thru 0. 

If R = Q and M = 0, the given forces balance and the body acted 
upon is in equilibrium. 

The above three methods of solving problems of co-planar forces 
should be well mastered. The student will see that each method 
has at times special advantages. 

89. IV. The funicular, or chain polygon. Definition: A chain 

polygon is a device for finding graphically a point in the resultant 
of a system of co-planar 
forces acting at differ- 
ent points on a solid 
body or frame, by the 
introduction of a series 
of internal forces which 
successively balance 
the given forces, and 
which are themselves 
balanced by two forces 
whose lines of action 
intersect and thus determine a point in the resultant required. The full 
meaning of this definition will be seen if a solution is closely followed. 

1. Let a trapezoidal frame, consist- 
ing of seven bars with pin-joints, be 
acted upon, as shown in Fig. 92(a), by 
five co-planar forces Fi . . . . . F$. 

The resultant of the system is to be 
found. 

The magnitudes and directions are 
shown in (6), as in the last section. 
The open force polygon is BCDEHK, 
and the Resultant, BK, is found in 
magnitude and direction. 

2. We are now to find its position. 
From any convenient point on the 

right of the force polygon, Fig. 926, draw right lines to B, C, D, E, H 
and K. These lines represent tensions in the links of our proposed 





Fig. 92 



70 NON-CONVERGING FORCES IN A PLANE 

chain. The force, Fi, acts between the two links whose tensions are 
BO and CO. The force lines of F u and the two tensions form a static 
triangle for the pin at j)\, which point is taken at random on the action 
line of Fi, in Fig. a. If parallel to BO, we draw bo so as to intersect 
the line of action of Fi, and then thru the point of intersection draw 
co parallel to CO, we have the lines of action of the three forces (viz. : 
Fi and the first and the second links of the chain) which balance. 

Producing co till it intersects the line of action of F 2 , and drawing 
do parallel to DO, we have the action lines of the forces whose static 
triangle is COD. In the same way drawing do, eo, fo and ko, we have 
all the forces balanced except bo and ko. The tensions in bo and ko 
must balance them all, and this is shown by the static triangle BOK, 
and hence bo and ko must balance the resultant. It follows that the 
point where they intersect, Q, is in the resultant R. Thru Q, parallel 
and equal to BK, the resultant of Fi, F 2 , F 3 , F 4 , F 5 can now be drawn. 

3. The point is called the "Pole"; the lines meeting at the pole 
are called "rays"; Fig. 92 (a) is called the "space diagram"; Fig. 92(6), 
is the force diagram; and the polygon in (a) formed by the interior 
links and the lines PiQ and P^Q, is called the Equilibrium Polygon. 

90. Different chains. The student can choose a second "Pole" 
for 0, and so have a different chain, and find a different Q, but if the 
figure is carefully drawn the new point will still be on R. 

Great freedom is allowed in drawing a chain polygon just as there 
is in a force polygon; the forces may be taken in any order, and it 
is quite possible that one or more links would be in compression, 
i. e., be struts instead of ties. 

1. When the given forces are all vertical and acting down, the 
force diagram should take the forces in order from left to right, and 
the pole should be on the right, if the 



5# 



<# 



20# 



Z8# 



«# 



links of the chain are to be in tension. 

If the pole be taken on the left, or if 

the forces be taken in the reverse order, 

the links will be struts and they will form 

an arch. It is not necessary that the points P u P 2 , etc., are actually 

on the body. Since the shape and extent of the body are indefinite, 

the pins of the imaginary chain may be off or on. 

Examples. 

Ex. 1. Find the chain (made of tie bars and pins) which will 
balance the above forces, assuming that the end links are anchored. 

Ex. 2. Find an arch which will balance the above forces acting 
on a body, assuming that the end struts have suitable supports. 



WHEN THE RESULTANT IS A COUPLE 



71 




Fig. 94 a 



Fig. 94 



2. If the force polygon closes, thereby showing that the result- 
ant is not a single force, the first and last rays coincide, and the first 
and last links in 
the chain will 
represent equal, 
opposite and 
parallel forces, 
thus forming a 
balancing couple. 
The system 
shown in Fig. 94 
gives a resultant couple whose moment is balanced by Pip, see Fig. (a). 

3. Should the lines of action of Pi and P 5 , in Fig. (a), coincide, there 
would be no couple, and as both R = and M = 0, and the given forces 
would be completely self-balanced. 

91. The closing link and vertical supports. Having found the 
resultant of a system of forces by the IV. method, it is often neces- 
sary to resolve it into parallel components acting thru given points. 
This is readily done by the device of inserting an additional link connect- 
ing the first and last links of the chain, or the first and last bars of an 
arch. The stress of this new link combined with the old terminal links 
separately gives the parallel components of the resultant. This method 
will be illustrated. 

Let the given forces be vertical, and let one of the required parallel com- 
ponents, which will, of course, be vertical, pass thru the point A. Fig. 95. 

R Drawthe 



1 


o 


o 


o 




F, 




;- 
























Fl 




e _____ 




h 








*i 


6^ 






F t 


1 




' 




/k 


v 2 


1 


_ c 


Vjt 








0^- 


■^/ D 








*2 




- 


C 


/ 














a 


\ 


R 










V 






















1 


% 



Fig. 95 



force polygon 
BCDK, with 
the resultant 
BK. 

Choosing 
now a pole for 
the stresses in 
the chain, 
draw the 
"rays" and 
construct the chain polygon as before, starting at A, and find the position 
of the resultant R. 

Now, let us suppose that the other parallel component of R is to 
pass thru the point E, in the last link of the chain. We then insert 
a new link from A to E, and draw the "ray" parallel to it from in 
the force diagram. The length eo gives the stress of this link and the 
two components of R, which are BE and EK in the stress diagram, 




72 



NON-CONVERGENT FORCES IN A PLANE 



are determined. If these components are individually .supported 
and balanced by direct forces, V\ and I 7 2, we have parallel supports 
which would completely balance a given system of forces. 

If now we consider V\ and Vi as additional forces acting upon the 
given body, we have a set of vertical forces which balance, and the 
complete closed chain forms the Equilibrium Polygon. It is, in 
fact, the "static polygon" of the chain. 

The reader should take note of the static triangles of the several 
points or pins, A, B, 0, D and E, in Fig. 95(6). 

92. When both vertical and inclined forces act en a structure, 
as in the case of vertical loads and wind pressures on a roof, the re- 
sultant is in general not vertical. If now one of the supports is to 
be vertical, the other must be inclined, and the resultant must be 
resolved into two components, one of which is vertical and acting 
from the point where the support is to be vertical, the other inclined 
and acting through the point where the inclined support must act. 

In such a case, the line of the resultant already found, should be 
produced until it intersects the vertical support-line; a line drawn from 

their intersection to the other support will 
give the direction of the inclined support. 

Suppose the resultant of a system of 
vertical and inclined forces, acting on a roof, 
has been found, and is represented by R. 
Fig. 96. The two supports are at A and E. 
The roof is anchored at E, and rests on rollers 
at A, thereby making the support Vi at A 
vertical. 

The above furnishes a method of finding 
the supports for the problems solved in 
Graphical Statics, Chapter XI. 

93. If both supports are to be inclined, any point in the line of 
action of the resultant may be taken as the point from which the two 
supporting action-lines shall 
pass, and their inclinations may 
be chosen at will. In such a case 
the supports, V\ and F 2 , are 
readily found. 




Fife. 96 




Fig. 97 



94. An equilibriated arch. 

In the case of an equilibriated 

arch, the supports are always 

inclined as illustrated below where the given forces are all vertical. 

Fig. 97. 



FORCES IN SPACE 73 

The student who later makes a special study of arches will find 
another and most important use of the equilibriated arch. 

Problem. 

Let the student assume a series of equal vertical loads, placed at 
equal intervals, and support them by an equilibriated arch; and again 
by a chain; and see how closely the balanced pins lie in parabolic 
arcs. 



CHAPTER VI. 

Forces in Space. 



95. Case I. When three forces meet at a point. Graphical. When 
three forces not in the same plane meet at a point, they cannot possibly 
balance; they can, however, be balanced by a fourth force acting 
thru the common point. The fourth force must be equal and directly 
opposite to the resultant of the three. Hence, to find the equili- 
briating force, we first find the resultant of the given forces. The 
resultant can be found graphically by drawing in different planes. 
Thus the resultant of two can be combined with the third in a new 
plane. The resultant vector of three forces is evidently the diagonal 
of a parallelopiped. 

Pi, Fig. 98, is the resultant of Pi and P 2 ; and the resultant of Pi 
and P 3 is P, which is seen to be the diagonal 
of a parallelopiped, which has the given force 
lines as diverging (or converging) edges. The 
drawing, however, is not a scale drawing, and 
has little value beyond giving clear mental 
concepts. When four forces balance, we have 
a static polygon (putting P 4 = — R,),OABC — 0, 
which is a warped static polygon. 

96. Analytic. 1. The only convenient method of recording 
the data and reaching exact solutions of such space problems 
is the analytic, using three rectangular axes, and specifying direc- 
tions by the angles which force lines make with them separately. 
We shall, in general, assume that OX and OY are horizontal, and that 
OZ is vertical. The logic of the arrangement has already been ex- 
plained. 

2. The angles which a force P, makes with the axes 0X 9 OY and 
OZ, are respectively a, /3, y, the arrows always pointing from the origin. 




CONVERGING FORCES IN SPACE 



It is evident that these angles are not wholly independent of each 
other. The angles lie in different planes, no one of which (except in 
special cases) is a co-ordinate plane. 

Suppose we know, at first, only that a is 40°. At once we picture 
a cone of revolution with for its vertex, OX for its axis, and a half- 
vertical angle of 40°. If now we learn that /3 is 60°, we think of another 
cone mounted on OY with a half vertical angle of 60°. Of course, 
these cones intersect in two elements, one above and one below the 
plane of XY. There are therefore two possible values of y which 
are supplements of each other, which are both 
determined, and P is either in the first or fifth 
triedral angle. 

3. The trigonometric equation showing the rela- 
tion of the three angles is found as follows: 

If the force P be resolved into its rectangular 
components Pi along OX y P 2 along OY, and P 3 
along 0Z y we have by inspecting Fig. 99: 

Pi = P cos a 
P 2 = Pcos/3 
P 3 = Pcosy 

Squaring and adding, and taking note of the fact that 
Pi 2 + P 2 2 + P 3 2 = P 2 , we have 
cos 2 a-f cos 2 /3+cos 2 y= 1; (l) 

a = 40° and /3 = 60° 




1 ig. 99 



hence, when 



cosy= ± ■v / (l-cos 2 40°-cos 2 60°) 



giving the two values of y already noted. 

4. When the resultant of several forces acting at the same point 
in space is desired, for the purpose of finding the balancing force, 
proceed as follows: 

(a) Assume the common point of action as the origin, and if 
possible, take the co-ordinate planes so that at least one of the forces 
will lie in a co-ordinate plane. 

(b) Ascertain the direction angles of every given force. 

(c) Resolve every given force into its three rectangular com- 
ponents. 

(d) Add the components on the separate axes, getting 

Pi = SPcosa j 

P 2 = 2Pcos/3l (2) 

P 3 = 2Pcosy ) 



A PRACTICAL PROBLEM 



75 



(e) Find R by the equation 



r= tJrs+rj+r? 

and a r ,/3 r and y r by the equations 

cos a r = Ri/R 

cos/3 r = R2/R ( 3 ) 

cos y r = RzlR 

97. Illustrative example. A weight of 100 lbs. is to be sus- 
pended to a small ring directly over it, and that ring is to be held in 
position by three 
cords. The posi- 
tions and tensions 
of two of the cords 
are known, but 
nothing is known 
of the third cord 
and force except 
that it must bal- 
ance the three 
others. The un- 
known force is to be 
fully determined. 

The positions 
and magnitudes 
of the given forces 
are shown in Fig. 
100. 

Solution. Take 
the ring as the 
origin, and the plane of Wi and W 2 as the plane XZ. A horizontal 
gridiron frame, three feet (3') above the ring, supports light pulleys, 
over which cords are to run from the weights PT 2 = 80 lbs., and from 
fl7 3 = 40 lbs. The tension line of W 2 is thru A, which is in the plane 
Z% and 4' from 0'; hence 

cos ct2 = - ; cosy 2 = — ; cosp 2 = 
5 5 ' 

The tension line of JF 3 = 40 lbs. is in a vertical plane which bisects 
the angle (+F)0(-X), and the pulley B' is 5' from 0', and 3' above 




Fig. 100 



the plane XY; hence tany 3 = 



and cos y 3 = — = 
V34 



76 



CONVERGING FORCES IN SPACE 



To find /_? 3 , assume a spherical surface, with radius unity and center 
at 0, cutting OF at iC, and the force line of W 3 at H. 

The bisecting plane, passing thru B r and OZ, cuts the sphere in 
the arc HN, which is the complement of y 3 . Drawing the arc NK 
cut from the unit sphere by the plane XY, we have a spherical right 
triangle HNK, right-angled at N. Its legs are KN = 45°, HN = 90 - y 3 , 
while HK = &. 



Hence, by Napier's Rules, 
and by 96 (1) 



cos/_?3 = cos 45° sin y 3 = — Vl7 
34 



cos 



/ , 9 25 
a 3 = — \ 1— - — -— ■ = 
\ 34 68 



34 



Vl7. 



The reason for the negative value of cosa 3 is evident. 
Table giving data and solution: 



p 


cos a 


cos /3 cosy 


P cos a 


P cos ft 



Pcosy 


JFi^lOO 








-1 





-100 


W 2 = 80 


0.800 





0.600 


64.00 





48.00 


W 3 = 40 


-0.606 


+ 0.606 


0.515 


-24.24 


+ 24.24 


20.60 


P=? 


a r 


& 


7r 


Ri = 39.76 


R 2 = 24.24 


Rz= -31.40 



R= V#i 2 + # 2 2 + #3 2 = 56.16 

The position of R is seen from its three components, R lt R 2 and R 3y 
in Fig. 100, to be in the 5th triedral angle. 

The angles a r , /3 r , y r are found from their cosines, which are 

39.76 24.24 -31.4 

cos dr = - — , cos p r = —-^-^ , cos y r = - 



56.16 



56.16 



56.16 



If we represent the direction angles of the balancing force by 

a' = 180 o -a r , 
$' = 180° -&, 
r ' = 180°-y r , 
we have . 39.76 



cosa = — cos a r = — 
cos/3'= —cos/3 r = — 
cos y ; = — cos y r = + 



56.16 

24.24 

56.16 

31.4 



56.16 



and the Balancing Force is completely found. 



FORCES AT DIFFERENT POINTS IN SPACE 



77 



F 


a 


<90° 


r 


20 


60° 


50° 


85 


>90° 


30° 


90° 


100 


135 


100° 


>90 


60 


30° 


120° 


? 


R=? 


a r = ? 


k-' 


7r=? 



To exhibit the position of its line of action more clearly, it can 
be said that the force required to balance the three given forces acts 
up into the third triedral angle. It pierces the plane X'O'Y' (which 
is three feet above XO Y) at a point P, which is : 

Stany' distant from 0', 

3 secy' cos a' distant from Q'Y\ 

3secy'cos/3' distant from O'X'. 



Problem. 

Let the student mentally picture 
the concrete situation whose data 
are given in this table, and repre- 
sent them graphically; then derive 
the answers to the five questions. One 
square in the table is purposely left 
vacant. 



Queries. 

1. What does it mean when one of the components of R is zero? 

2. What does it mean when two of the components of R are zero? 

3. What does it mean when all three components of R are zero. 

98. Case II. When the forces do not converge to a point, 
but act at different points and in different directions. The General 
Case. The analysis is an extension of the analysis for non- 
concurrent forces in a plane. A point of application is given by its 
three co-ordinates, x, y and z. The direction of a force by three 
direction angles, a, /3, y, with reference to three rectangular lines 
(parallel to the co-ordinate axes) drawn thru the point of application. 
The steps in the process of finding a resultant are as follows : 

(a) Resolve every force into its three components : 

Pi = Pcos a 
P 2 = Pcos/3 
P 3 = Pcos y 

(6) Resolve every component into an equal parallel force at 0, and two 
moments, with reference to the two co-ordinate axes perpendicular to 
it. This must be illustrated. Let Pi, Fig. 101, the component acting 
at the point A(x, y, z) be resolved into an equal force, P/, acting B, 
and a couple zPi {right-handed) about the Y axis. Next think of P/ 
as acting at C, and resolve it into an equal parallel force, Pi", at 0, 
and a left-handed couple yP\ about the axis Z. Of course Pi produces 
no moment about the X axis, with which it is parallel. 



78 



RESULTANT OF FORCES IN SPACE 



We thus get for Pi acting in space: 

a force Pi at 0, along OX: 
a moment about OY of -\-zP\\ 
and a moment about OZ of — yP\. 

In like manner, for the component P2, we shall get: 
a force P 2 at 0, along OY; 
a moment about OZ of +.tP 2 ; 
and a moment about OX of — zP 2 . 

And for the component P 3 : 

a force P 3 at 0, along OZ; 
a moment about OX of -\-yP$; 
and a moment about OY of — #P 3 . 

Summing these results, we get for every force P in space: 
Three components, Pi, P 2 , P3, at the origin of co-ordinates; 

A moment about OX of yP 3 — zP 2 = M x 

A moment about OY of zPi — xP 3 = M y . 

A moment about OZ of #P 2 — yP\ = M z . 

A careful examination of Fig. 101 
will make all these moments intelligible. 

If the results are tabulated the 
student will notice the sequence of 
+ x letters and numbers. 



+-Y C 



±z 






} 






P* 


A 

v 


p^ 


1 


, 


p l 


/ 


N pin / 
' / 




} X 




/y 




z 


/v 


/Pi , 











Fig. 101 



p 


M x 


My 


M z 


Plp 2 p 3 \yp3-zp2 


zPi-xPs 


xP 2 -yPi 



Note the sequence of letters and numbers, in Fig. 102. 

The student will also notice that the moment about 
0Z or M 3 is identical with that found for the axis OZ, 
when the force was in the plane XF, and there could be no 
moment about OX and Y as both z and P 3 were zero. 

(c) After all the forces are resolved as above, sum the 

results. 

Find 2 (Pi) =R X 2 (yP 3 ) - 2 (>P 2 ) = M 1 

Find 2 (P 2 ) = R 2 and 2 (*Pi) - 2 (*P 8 ) = i/ 2 /■ 

Find 2 (P 3 ) = P 3 2 0P 2 ) - 2 (2/P1) = M 3 ) 




Fig 102 



(d) 



fi= VW+i^+fls 2 ) 



i/= V(J/ 1 2 +M 2 2 +M 3 2 ) 



(4) 



(5) 



SOLUTION OF A GENERAL PROBLEM 



79 



This combination of three moments about three separate rectangu- 
lar axes is in accord with what was shown in (II). 

In general, the resultant of such a system of forces in space is a result- 
ant force acting at the origin, and a resultant moment about an axis in space. 
(e) The direction angles for R are found by the equations 

cos a r = Ri/R *) 

cos /3 r = R 2 /Rr (6) 

cos y r = R 3 /R J 
of the axis of the resultant moment are 



And the direction angles 
found by the equations 



(7) 



cos X = Mi/M 

cos fi = M 2 /M 

cos v = M s /M 

2. The axis of M may be assumed to pass thru 0. If the axis of M 

is parallel to, or coincides with, R, no further combination is possible. 

If the axis of M and R make an angle 0, M may be resolved into 

two new components, M cos 0, and M sin 0. The component, Msin0, 

can be combined with R> thereby shifting R to a parallel position 

distant from by a perpendicular L. 

Msin0 



L = 



R 



The perpendicular is drawn normal to a plane which passes thru R 
and is parallel to the axis of M . The resultant of the system is then 
R in this new position and the moment Mcos0. If is a right angle, 
Mcos0 is zero, and the resultant is a single force. 

If M = 0, the resultant is a force R at the origin of co-ordinates. 
If R = while M is not zero, the resultant is a couple with moment M. 

If R = and M = 0, the system is self-balanced and the body acted 
upon is in equilibrium. 

While an engineer rarely has occasion to solve a general problem in 

statics, it is well to embody the method in the solution of a problem 

with assumed data. t. , . 

Problem. 

Solve the problem in the table, filling out blank spaces systematically. 



p 


X 


y 


z 


a 





r 


Pi 


P 2 


Ps 


yPs 


ZP2 


zPi 


xP 3 


xP 2 


yPi 


8 
16~ 


4 


5 


12 


75 


50 


<90 c 












— 




O 








30 


60 


60 












10 


—2 


+3 


—4 


135° 


>90 








20 


8 


— 1 


—6 


>90 


70° 


45 c 




























Rv=? 


R 2 =? 


Rs-? 




















R=? 




M=? 










Mi=? 


M 2 =? 


M 3 =? 



80 CENTROIDS OF PLANE SURFACES 



The student will notice, when P = 8, that cosy =+ Vsin 2 a — cos 2 /8, 
since y must be acute. 

The actual value of y need not be found, as only its cosine is wanted. 

The value of (3 for the second force the student must supply himself. 

The above system of forces can be balanced only by balancing R, 
and il/, either by a single force or a combination of a single force and 
a single couple. 



CHAPTER VII. 

Centroids and Centers of Pressure. 

99. 1. The two most important applications of the theory of 
parallel forces are found in problems which require the concentration 
of distributed forces. In Chapter I it was assumed that such con- 
centration was possible, and we did not hesitate to give the total 
magnitude of a force distributed over a surface or thru a volume; 
but we did not attempt to determine the exact location of the line 
of action of the concentrated force. This chapter will be devoted to 
finding the positions of concentrated or resultant forces when the law 
of distribution is definitely known. 

100. The point in a surface where the resultant of a distributed 
force pierces that surface, is called the "Center of Action." 

As has been said, all of our problems are more or less ideal, but they 
closely approximate the real. We assume, for example, that the 
steam or air pressure upon the vertical face of a piston is uniform, 
tho we know that such is not absolutely the case. We assume that 
the pressure between the corner stone of a building and the material 
below it is uniformly distributed, when it probably is not uniform. 
We assume that the earth's attraction upon a cubic inch of material 
in a block of wood, stone or metal, is uniform thruout the block, 
tho we have no idea that such is the absolute truth. 

101. The phrase: "The intensity of the surface action at a 
point* ' means : the amount of action or force there would be on a unit 
of surface if thruout the unit surface the force were uniformly what 
it is upon the infinitesimal area surrounding the given point. 

Since the magnitude and direction of a uniformly distributed force 
cannot affect the position of the Center of Action, we may assume 
that the direction is normal, and that the intensity is p, which is as 
small as one likes. 



THE CENTROID OF A TRAPEZOID 81 

102. Centroids of plane surfaces. The point where the resultant 
of a uniformly distributed force pierces a plane surface is called the 
"Centroid"* of that surface. 

In some simple geometrical surfaces the centroid is evident by- 
symmetry, without demonstration. For example: The circle, the 
ellipse, a rectangle, all regular polygons. If a surface has an axis 
of symmetry, the centroid is on that axis; two such axes determine the 
centroid. If a surface can be divided into parts such that the cen- 
troid of every part is known, and known to lie on a certain straight 
line, the centroid of the surface is known to lie on that line. The 
number of parts may be finite or infinite. Hence the centroid of a 
triangle is at the intersection of its medial lines. 

10«$. When a surface of action can be divided into finite parts 
whose several areas and centroids are known, the problem of finding 
the centroid of the entire surface falls directly under parallel forces 
in space. Chapter III. 

It was proved in geometry that the point which we now call the 
centroid of a plane triangle is on a medial line two-thirds of its length 
from the vertex end. The centroid of 
a plane trapezoid is found graphically 
as shown in Fig. 103. The diagonal 
AB divides the surface into two tri- 
angles whose centroids determine the 
right line, ab. The diagonal CD, gives 

-i 1 it Fig. 103 

two new triangles, and a second line, 

cd, which, with ab determines G, the centroid of the trapezoid. 

The centroid of any rectilinear plane surface is found by an exten- 
sion of the method just used. 

104. The general case. When the outlines of a surface are 
curved and irregular, and there are no axes of symmetry, the Centroid 
must be found by the doctrine of moments and the use of the Calculus. 

1. Let the uniform force intensity be p. If A represents the whole 
area of the surface of action, A A will represent a portion of that sur- 
face, and p A A will represent the magnitude of the force on that por- 
tion. As the whole is equal to the sum of its parts 
R = H(pAA) 

* The author prefers the word ' 'Centroid' ' to the phrase "Center of Gravity" 
when speaking of plane areas, yet the latter is in very common use. 

The introduction of thin flat plates of heavy material fails to hide the absurd- 
ity of speaking of the gravity of a surface; and to go further and assume that 
the plate is half above and half below the given area in order to bring the C. G. 
of the plate into coincidence with the Centroid of the surface is confusing to a 
novice. 




82 



THE CENTROIDS OF PLANE SURFACES 



In this case the summation must be made by the Calculus. Instead 
of the sum of a finite number of finite terms, we must have the sum 
of an infinite number of infinitesimal terms. Hence we shall use 
for 2, and d for A, and the above formula becomes 



J 



R = fpdA 




Fig. 104 



It is here and elsewhere assumed that the student is familiar with the 
ordinary methods of integration, but he may be less familiar with 
their application to practical problems. Accordingly, some funda- 
mental matters will be fully explained. The 
expression dA is the numerical measure of a 
portion of the surface A, and preferably repre- 
sents the product of a length by a breadth. 
Both factors may be infinitesimals, as dxdy; or 
one may be finite, as ydx, or (y 2 — yi)dx; see 
Figs. 104 and 105. If dA=dxdy y two integra- 
tions will be required; if dA=ydx, only one 
integration is necessary; hence the second form is preferred, when 
it can be used. 

2. Let BC he sl surface bounded by outlines not at present defined. 
General formulas for the position of its centroid are to be found. 
Let the surface A be divided into elements parallel to the axis OY, 
with a constant width, dx. The elementary area of a single strip is 
(yi — yi)dx 9 according to the lettered Fig. 104. If p be the intensity 
of the distributed force, the force acting upon the strip is (y 2 — yi)pdx. 
The moment of this force about OY is x(y 2 — y 1 )pdx i and the sum of 
the moments of all the elements is 

X1 x(y 2 -yi)pdx = pj xi (y 2 -yi)xdx (l) 

since p is constant during the integration. 

pa® 
Now, the total force acting on the surface, BC, is p I (y 2 — y{)dx 

which is p times the area, or pA. When this force is centralized it will 
act at the centroid of the surface whose co-ordinates we will call 
x and y . Then the moment of the centralized or resultant force 

/»X2 

about OY will be % V J x (y2 — yi)dx = M y 
Now follows an Axiom of Mechanics: 
The moment of the whole is equal to the whole of the moments. 



J i% X 2 /*X2 

X1 (y2~yi)dx = p j xi (y 2 -yi)xdx 



xM = M„ 



GENERAL FORMULAS 



83 



Hence 



x n = 



) (V2-yi)xdx 
(y 2 -yi)dx 

J x\ 



R 



(2) 



3. The method of integration is determined in every case by the 
equations of the boundary lines of the surface. 

The disappearance of p shows that x depends not on the absolute 
value of p, but upon its uniformity. We will therefore assume that 
p = \, and numerically R = A. 

4. The co-ordinate, y , is found in a similar manner. If the same 
elementary strip, (y 2 — yi)dx, is used, we must get its moment about 
OX by multiplying its area by the ordinate to its central point, which 



is 



2/1 + 2/2 

2 



Hence our value of y will appear 

i c x2 

2 J X1 (y 2 2 -yi 2 )dx 



Vo 






y\)dx 



R 



Mi 

A 



(3) 



5. If a new element is taken parallel to the axis of X, the value of 
y appears thus: — s* y 

\x 2 - xi)ydy 

y = 



M, 



I {x 2 ~xi)dy 



(4) 



There are other general methods depending upon the form of the 
surface element and upon the use of circular or polar co-ordinates. 
A variety of illustrative examples is chosen so as to make every method 
clear. 

105. Examples illustrating methods of finding centroids. 

Ex. 1. The centroid of a parabolic half seg- 
ment. Fig. 105. 

Let the surface be OAC, and the element be 
PB = ydx. The Equation of the curve is y 2 = ax. 

y , the a 



N. B. — While we are to find 



and 



dx 
Fig. 105 



co-ordinates of the interior point G, the student 

must bear in mind that x and y thruout our analysis 

are the co-ordinates of points P, on the bounding curve. 

Since p = l, the force on our element is ydx, and its moment about 

the axis OF is 7 ,, , 

dM y = xydx 



Hence 



M 



= I a*x*dx = 



h 2 h 



84 CENTROIDS OF PLANE SURFACES 

It was found in the calculus that 

Area of OAC= - kh = A = R. 
3 

Hence by 104, x = ^ = - h. 

A 5 

The centroid of the strip element, BP, is at its middle point y/% from 
OX; hence, the moment of the element about OX is 

dM„= ydx 
x 2 J 

and M T = - I y 2 dx = - I xdx = 

2 Jo Oo 4 

y ° A 8 

106. The student must not be wedded to one form of a surface 
element. The element QD = (k — x)dy, would have served equally 
well. We see at once using subscripts 1 and 2 in place of x and y: 

dMi = {k — x)ydy 
Mi = J o kydy - • J o xydy 



M 2 _ A 4 M 2 






2 4a 4 






^ = *h 






A 8 






(A: — x) ay 






I ( (k*-x*)dy=^ - 


h 5 
10a 2 


= -k% 
5 


^ = S L 






A 5 







dM 2 = (k 
M 



lOT. The algebraic method. This method is based upon a 
full set of data, and is best illustrated by an example. Suppose we 
wish to know the centroid of the parabolic spandril OQAE, knowing 
the centroid of the rectangle OCAE, and the centroid of the half 
segment OCA, as found above. 

The three centroids are on the same straight line. Why? And 
the centroid of the spandril is twice as far from the center of the rect- 
angle as is the G of the segment. Why? 



ANALYTICAL METHODS 



85 



The Axiom of Moments with respect to each axis, OX and OY, 
serves our purpose. Letting x o9 y be the unknown co-ordinates of 
the centroid of the spandril, and using the data of the rectangle and 
segment, we have the two moment equations. 

M 2 = hkX-k = -hkX-k+-hkXx 
2 3 5 3 



Mi = hkX-h 

2 



2 3 1 

-hkX-h+-hkXy 

3 8 3 



whence 



x=—k 



Vo 



10 
3 



h. 



4 



This indirect method is often very convenient. 

Ex. 1. Find the Centroid of the above Spandril by direct integra- 
tion. 

Ex. 2. Find the Centroid of a quadrant of an ellipse. 

Ex. 3. In Fig. 103, let the non-parallel sides, AC and DB, be pro- 
duced till they meet in 0. Find graphically the centroid 
of the triangles OAD and OCB, and measure their 
distance from 0. Then by the algebraic method find 
the distance to the centroid of the trapezoid. 

108. Find the centroid of a circular segment, intro- 
ducing circular co-ordinates. 

Solution. The segment is ABC, Fig. 106. Since 
OX is an axis of symmetry, the centroid is on that 
axis and y = 0, The element is %ydx. From the figure 

x = rcos0, y = rsm0, dx=— rsinOdO, xi = rcos(f> 




M y = I lyxdx = — 2 I _ r 3 sin 2 cos 6 dO = 2r 3 I i 



sin 2 # cos dO, 



sin 3 (/>. 



A = % \ydx= —2r 2 j sin 2 6 d0 = %r 2 I sin 2 d0=r 2 ((j>— sin (£ cos </>) . 

The student will take note that the expression for A (the area of 
the segment) is explicitly the area of the sector OACB minus the 
area of the triangle OAB: that is 

A=r 2 (j) — r 2 sin (j) cos cj>. 
Dividing M y by A we have 



M. 



x n = 



3 '(p 



sin 3 (f) 



sin<£cos<^> 



86 



CENTROIDS OF PLANE SURFACES 



For a semi-circle <f> = 7r/2 and 

4r 

x = — • 

3?T 

Note. The student must not be surprised that dx = — r cos 6 dO 
with the negative sign. If he looks at Fig. 106, he will see that as 
x increases, 9 decreases; hence, if dx is positive, dO must be negative, 
and ( — d9) is positive. 

109. The centroid of a circular sector. The surface element may 
best be defined by polar co-ordinates. Let the given surface be the 

sector of a circle, OABC, Fig. 107. As OB is 
an axis of symmetry, we have only to find the 
distance x . This could be found by algebra if 
we knew the position of the centroid of the cir- 
cular segment ACB, as well as of the triangle 
OAC. But the direct method is simpler, as 
follows : 

Let the radius be r, and the half-angle 
A0B = (f). Draw two consecutive radii in a general position defined by 
the arcs and dO; the elementary surface is an infinitesimal triangle 




A d0\ 



whose area is 
distance from 
integrating, we have 



2 



its centroid is at a distance — r from 

3 



YY is - rcosO. 
3 



0, and its 
Hence, taking moments about YY, and 



x = 

o 



Pr: 

Jo 2 



rcos 



OdO 



2 



X 



<td 



sin<£ r 

T 



Both integrals are doubled so as to include the whole sector. 
Corollary. 

If 6 = 7r/2, the sector is a semicircle and x Q = 4/7377 
as found in 108. 

Ex. Find the centroid of the "spandril" to a 
segmental arc. 

HO. Double Integration. To find the cen- 
troid of a sector of a circular ring area, EABCDN. 

Fig. 108. We now take an elementary part of such 
a sector enclosed by two cosecutive arcs, and two 
consecutive radii. This element being an infinites- 
imal of the second order is treated as a rectangle 
with the area (pd0) (dp). Its moment about YY is p 2 dpcos0d0. 




Fig. 108 



DOUBLE INTEGRATION 87 



Hence 



l XJ P 2d pcos0d0 
J r ^fjpdpd0 



The order of integration is immaterial. If we integrate for 6 first, 
we shall get the moment of a sector of an elementary ring as shown 
by adjacent arcs; its length is 2pcf> and its width dp. The next integ- 
ration would sweep the ring outward from p = n to p = r 2 . 

Had we integrated first for p, we should have had a trapezoid lying 
between two consecutive radii and extending from r± to r 2 . 

Taking the first order, integrating for 6 while p and dp are constant, 
we get (canceling the 2) 



sine/) jr'p'dp g sin( £ r2 s 



r P d P 



<t> r 2 



2_ ri 2 



This problem could have been done by the algebraic method and 
109. 

If ri = 0, we get the centroid of a sector as in the 
last example. 

111. The centroid of a circular arc. Fig. 109. 
The arc may be supposed to have a breadth of dr, 
the radius being r. As above, the element is 
rdrdO; its moment* about YY is (rdrdO) (rcos9). 

M y = 2r 2 dr I cos 0dO = 2r 2 sin <j)dr 

The area of the arc, or infinitely narrow ring is 2rcf)dr. 

_ M y _ 2r 2 sin</> _ rsin<^> 
A 2r(f> </> 




(10) 



Another form of this result is easily remembered : Since the chord 
A C = %r sin (f), and the arc =%r(j), we have 

x _ 2rsin(/> _ chord 
r %r(j) arc 

which proportion is readily put into words. 



* Since we have obliterated the force, p, per square unit of surface, by mak- 
ing it unity, we naturally drop into the habit of speaking of a surface as tho it 
were a normal force. The ''habit" is quite conventional. 



88 



CENTROIDS OF PLANE SURFACES 



Problem. What is the length of the arc whose centroid is at the 
center of its central radius? 

112. The centroid of a cycloidal area. Suppose the area under 

consideration is bounded by 
a cycloidal arc and its base. 
Fig. 110. It is evident that 
BH, the longest ordinate, is 
an axis of symmetry: hence, 
x = 77Y, and only y is to be 
found. 

From the figure, since ON = rO 

x = r0 — rsm 
y = r — rcos0 

The element of the surface is ydx, and its centroid is distant from 
OX a length y/2. Hence ^ lTr 

y = 




Mi 

A 



Jo V dx 
From x = r0 — r sin#, we have 

dx = r(l -cos#) d6, so that 

„3 



Vo = 



f< 



(l-cos0) 3 d0 



%r° 



r<- 



Snr 2 



cos 



OydO 



The limits for follow from the facts that when x = 0, = 0; and 
when x = %TTr, = %tt. The uncanceled denominator incidentally 
shows that the area of the cycloid is just three times the area of the 
rolling circle. 

CENTERS OF ACTION OF DISTRIBUTED FORCES WHICH 
VARY IN INTENSITY. 

113. When a surface action is a distributed force which varies 
in intensity according to a definite law, the center of the force for 
the whole can be found if the centers of the parts can be found. The 
most important case is that of a uniformly varying force, acting on a 
plane surface. The following are cases where the variation is approxi- 
mately uniform: 



HYDROSTATIC PRESSURES 



89 



1. The pressure of water upon the wall of a tank. 

2. The normal force acting upon the cross-section of an elastic 
beam which is slightly bent. 

3. The pressure upon the top of a column which has a rigid cap 
supporting an eccentric load. 

4. The pressure on the material under a foundation stone which 
is eccentrically loaded. 

All of these cases will be discussed with more or less fullness. 

114. The center of hydrostatic pressure. The intensity of in- 
ternal pressure at any point in any liquid which is at rest in an open 
tank or vessel is equal to the weight of a column of liquid and the air 
above it, having for its base a unit area at the point, and reaching to 
the top of the atmosphere. The weight of such a column of air, with 
its base at sea level, is about 2,116 lbs per square foot. This pro- 
duces a pressure of 2,116 lbs. upon the surface of the level liquid, 
and by so much increases the internal pressure of the liquid at 
every point. The weight of a column of the liquid 
itself from the point under consideration to the sur- 
face must be added to get the intensity required. 
See Chapter under Buoyancy. 

If p be the required intensity at a depth of x 
below the surface of a liquid, let w be the specific 
weight, or the weight of a unit of volume, of the liquid, 
and p the atmosphere pressure on the surface of the water; then 

p = p +wx 

In point of fact, in problems in Applied Mechanics, the pressure 
p is more often balanced by an equal pressure on the other side of 
the wall or vessel, so that the pressure of still 
air can generally be left out of consideration. 
The increased pressure due to moving air, and the 
diminished pressure caused by rarefaction, require 
special examination. Accordingly, the hydrostatic 
pressure varies with the depth. 

115. To find the total pressure on a given 
vertical plane surface and the "center of pres- 
sure," we proceed as follows, assuming that the 
atmospheric pressure is self-balanced: — 

1. Take an element of the area of action, at a depth of x below 
the top of the liquid. Fig. 113. The intensity being the same at all 
points at that depth, let the element extend across the given surface 
horizontally, so that its area shall be {y^ — yijdx. The pressure at that 




Fig. Ill 



■ ~z?\ 

c — ~ ' 


■f-*?!-! — 


/ * 


KJ-! z 

|se'"~jl ] x 2 - 


U 7 \h' Vi 


\ y*-yi- j— u] 



Fig. 112 



00 



CENTERS OF PRESSURE 



depth is xw per unit of surface, and the pressure on the element is 
wx{y2 — yi)dx y and the total pressure on the entire area is 



= w jl!(y> 



yijxdx. 



(1) 



2. The Center of Pressure involves the moment. Take for the 
moment axis the line CC\ Fig. 112, tangent to the upper edge of the sur- 
face of action. Let x / = x — x 1 and dx = dx'. The moment of the 
pressure upon the element is 

dM c = w(y 2 — yi) (xi+x')x'dx' 
and M c = w §\y 2 - yi ) (x x x' +x' 2 )dx' 

and x = y c (2) 

The general formula for y is found, when needed, in a similar 
manner. 

1 1 6 • Let the surface of action A be a rectangle with vertical 
edges, so that y is constant, and equals b, Fig. 113, so that the area is 
A = bh. 

1. Applying 115 (1), we have 



P = bw. 



.>■■> 



2 _ 



*-l 2 , , X2 + X1 

— =oivfi 



wA 



X* + Xi 

2 



wA 



h + 2xi 



2 2 2 2 

i. e., the total pressure on the rectangle BCDE is equal to the area of 
the rectangle multiplied by the pressure at the "mean depth." This 
y result was almost self-evident, and the same for- 
mula must be true of a circle, viz.: the area of 
, the circle times the intensity of the pressure at 
the center. 

2. The moment about CCi of the liquid pressure 
on the elementary area bdx is bwxdxXx', as shown 
in Fig. 113. Since x = Xi-\-x\ and dx = dx\ we 
have for the whole moment 



C\ 1 E 


\h 


|X' 




A 

< _ . 



B 

Fig. 113 



M c = bw J 



bwh* 



and x = — c 
P 



(xix' ' -\-x' 2 )dx' = bw 
wAh 



V2 ' s) 



(2h + 3xi) 



6 



(211 + 3x0 



2/* + 3 t ri 



3 A + 2xi 

in which x is measured from the top of the rectangle, CC\ and X\ is the 
distance from CC\ to the top of the liquid. 



THE TRIANGLE AND THE CIRCLE 



91 



Corollary. If the area of action extends to the top of the liquid, 
Xi = o and we have 

x ° = s h - 

117. 1. Let the immersed area be the triangle BBD. See Fig. 114. 
With the lettering shown, we have 

xydx in which 

bx' 

y= — > x = Xi~\-x f > and dx = dx'; 
h 



Hence 



bw n 



(x!x'+x f2 )dx' 



bwh 



(3zi + 2/*) 




u Fig. 114 

2. The moment about CC of the pressure on the element is wxydx f Xx' 



and 



M c = — I ( Xl x' 2 +x' s )dx' 



wbh? 
12 



(4*i + 3/0 



and 



M _ h /3/*+4zA 

x is measured from CC. 

Corollary. If the vertex of the triangle reaches the top of the liquid, 

Xi = 0, and 3 7 

x n = -h. 



The center of pressure is shown in Fig. 114, 
provided the vertex D is at the surface of the 
liquid. 

118. Let the surface in the vertical wall 
of a tank be a circle. Fig. 115. Let x, x ly 
X2, and x' be defined as in the last section. 
In this case y = 2 ^lx'(%r — x'). As intimated 
in 116, the total pressure is equal to the 
area, multiplied by the mean depth. 




,,,';,;■/..■■ :A 



Fig. 115 



P = (7rr 2 ) • w> hence 

P = irwr 2 (*i+r) 
and M c = j(ydx') (wx)x f 



92 CENTERS OF ACTION 

Changing the variable to we have 

y = 2rsm0 
x' = r (1 —cosO) 
dx f = rs'm0d0 

x — X\-\-r (1 — cos#) 
and hence 

M c = 2wr s [z.f\m 2 0(l-cos0)d0+r £"sin 2 (1 -cos0) 2 d0 ] 



M = 2i 



£i( 



— sin#cos# 



(9 



■f 



2 3 

20 — sin2#cos2# 



16 



M 

) 



— sin cos 2sin 3 # 



and 



V 2 2 8 / V 4 / 

Jf c r 4.Ti + 5r 



Zi+r 



Corollary. If the circle touches the surface of the liquid, #i = 0, 



\^£-ps^- 






^^w * 







Fig. 116 



and 



so 



that P = 2wr 3 I cos 2 0sin0d0= — 



and 






x = -r 
4 

119. The pressure on a semi-circle. If 

the circle is half above the surface of the 
liquid, and only a semi-circle is under pres- 
sure, the problem is wholly new. 

With the notation shown in Fig. 116 

x = rsin0 
dx = rcos0d0 
y = %r cos0 

%wr 3 cos 3 



■— —wr° 



M y = 2wr* | sin 2 cos 2 0d0 = 2wr* 



'gfl-singflcosgfl^ 

16 ~ t 



M y = 



Trior* _irr l 
8 2 



irr 4, r l 



w • — =w 
4 



tt M 3 

Hence #„ = - — = — irr 

° P 16 



THE CENTER OF HYDROSTATIC PRESSURE 



93 



120. The student must not fail to see that the Center of a varying 
Pressure is a different point from the centroid of the surface. The 
" Center of Hydrostatic Pressure" is always below the centroid of the 
surface of action, but the distance below depends upon the depth 
of the liquid. The deeper the liquid, the less the distance between 
the two points. 

The results in all the above cases apply to surfaces in inclined planes 
which are immersed — if the axis of X is a line of greatest declivity 
in such a plane. Just as the value 
of w (the specific weight of the liquid) 
disappeared from the final formulas 
for vertical planes, so w sin <j) will dis- 
appear for the inclined plane. Fig. 117. 

The practical application of these 
formulas for hydrostatics is readily 
seen. If a gate or a valve in the ver- 
tical wall of a dam, flume or tank, is to be supported by a single prop, 
or a pair of shaft bearings, the prop or shaft must be so adjusted as 
to act normally at the center of pressure and not at the centroid of 
the surface. If the gate or valve is to revolve freely on a horizontal 
axis, i. e., so that the moment pressure will balance on the axis, that 
axis must pass thru the center of pressure. 

The problems of this chapter are by no means purely abstract, 
given simply as exercises. Some of the formulas giving the value 
of x will appear several times in later chapters. 




Fig. 117 



121 




The theorems of Pappus. The general expression for the 
volume generated by the revolution of a plane area 
about an axis in the plane, not intersecting the area 
revolved, is (Fig. 118,) 

V = ZirfxdA 



Fig. 118 



integrated between proper limits; dA is a differential 
of the given area taken parallel to the axis of revolution. The integral 



by the axioms of moments. 
Hence 



J xdA = x A 



V = %7TX A 



In other words the volume of a solid of revolution is equal to the 
area of the generating surface, multiplied by the path described by 
the centroid of that surface. This is known as the Theorem of 



Fig. 119 



94 THE THEOREMS OF PAPPUS 

Pappus. It may be used to find a volume when x is known, or, con- 
versely, to find x Q when V is known. For example, if we know that 

4 

the volume of a sphere is -7rr 3 , we can find the centroid of a semi- 

3 

circle by the equation 2 

4r 

II. 

The surface of a solid of revolution is found by multiplying the 
length of the generating curve by the length of the 

\ path of its centroid during the revolution. Fig. 119. 

o \ Since 8 = 277 J xdl, in which S is the surface of a solid 

of revolution, and I is the length of the generating line, 
co-planar with the axis, and not intersecting it; and 
since x l = J xdl we have 

S = 27TX l 

which is the above theorem. 

Problems. 

1. Find the surface generated by revolving the cycloid about its 
base. 

2. Find the surface and volume of a "torus" generated by revolv- 
ing a circle about an exterior line in its plane. 

3. Find the volume of the solid generated by 
revolving the spandril of a parabolic half-segment 
about the axis of the parabola. Fig. 120. 

4. Find the volume of an oblate spheroid whose 
polar diameter is 26, and whose major diameter ° l Fig . i 2 o 
is 2a. 

CHAPTER VIII. 

Centers of Gravity. 

122. In Chapter III we studied the relations of parallel forces, 
both co-planar and in space, without considering the source or 
nature of the forces. In this chapter we shall deal with parallel forces 
which are distributed through volumes, such as the earth's attraction 



CENTER OF GRAVITY FOUND BY EXPERIMENT 95 

upon all bodies at or near its surface. In Applied Mechanics, we 
have no occasion to discuss the problems of physical Astronomy. In 
everything relating to structures and mechanism, the lines of action 
of the earth's attraction are so nearly parallel that they are assumed 
to be parallel, and in all cases the arrows representing gravitational 
forces point vertically downward. 

The magnitude of the earth's attraction upon a body depends upon 
the mass of the body. Since such attraction is known as the weight* of 
the body, the weight is proportional to the mass. In a body of 
uniform density, the masses, and therefore the weights, of equal volumes 
are equal. 

Definition of center of gravity. The earth's attraction upon all of 
the minutest parts of a body constitutes an infinite number of constituent 
parallel forces, whose resultant is the weight of the entire body. 

It will soon be shown that there is within the enveloping surface 
of every body a certain point thru which the resultant always passes, 
no matter how much that body be turned over and about in space. 
That point is called the Center of Gravity, or the Center of Mass of 
the body.** This chapter will be devoted to finding the C. G. of 
typical solids, which closely approximate the forms of real or 
imaginary bodies. 

123. The center of gravity found by experiment. If a body can 
be suspended by a cord or wire so that the axis of the wire coincides 
with the resultant line of action of the weight which it balances, we have 
one line which must contain the C. G. of the body. If, then, the body 
can be suspended differently, giving a second line of action, the 
C. G. must be the point where the two lines intersect. This method 
is frequently used for finding the C. G. of thin plates of irregular 
shape; for example, a piece of zinc or drawing paper representing the 
cross-section of a steel rail. In such a case the C. G. of the thin solid 
is anent the "centroid" of the surface. 

124. General formulas. When the bounding surfaces of a solid 
are given in mathematical language, the co-ordinates of the C. G. 
are determined as , follows : 

1. Let the position of the body be given with reference to a set 



* Under the subject of Deviating Forces in Chap. XVI, it is shown that a com- 
ponent of the earth's attraction is employed in compelling the mass to revolve 
about the earth's polar axis, and that accordingly the observed weight of a body 
does not measure the entire attraction. But the weight is still proportional to the 
mass. 

** In the case of such bodies as are considered in this book, the center of gravity 
coincides with the center of mass. 




96 CENTERS OF GRAVITY 

of three rectangular axes. Fig. 121. An elementary portion of 
the body is represented by the rectangular solid 

dV = dxdydz. 

The weight of a unit volume of a homogeneous 

material is known as its "specific weight." If the 

specific weight of the material is w, the weight of 

dV is 

dW = wdxdydz = wdV 

Fig. 121 v 

If the plane XY is horizontal, the moment of the weight of the 
element with respect to the axis, OX, is 

dMi = wdV.y = wy dxdydz. 

ano. /*/*/* 

M\ = J J J wydxdydz 

The successive integrations are to be taken between the limiting 
surfaces of the solid. If w is constant, it comes before the integral 
signs. The entire weight of the body is 

W = wjffdV = wfff dxdydz. 

If the co-ordinates of the C. G. are x , y z > we have by the axiom 
of moments ^ 

which becomes, if w is constant, 

) ydV J ( j y dxdydz 
V ° = ' V = ff'fdxdydz. 

2. Similarly, finding moments about OF we get 

M 2 - 



° W 



and 



fxdV f f fxdxdydz 



V 



3. If now the body and the co-ordinate axes be so turned that the 
plane of XZ is horizontal, and the moments about OX are again con- 
sidered, we get r irr r r r 

fzdV ff f zdxdydz 

z ° = y ~ y 

and the C. G. is found. 

If w varies, it does not disappear from the formulas. It is generally 
assumed to be constant unless the law of its variation is definitely 
given. 



THE C. G. IS IN AN AXIS OF SYMMETRY 97 

125. Center of gravity by inspection. The center of gravity of 
a homogeneous body is evidently at its geometrical center, if it has 
one; like the center of a sphere; the center of a parallelopiped; and 
the center of a cylinder of revolution, thin, thick or solid, with parallel 
bases. It is further evident that if a cylinder or prism have parallel 
bases whose centroids are known, the center of gravity of the solid 
is at the central point of the right line connecting the centroids of the 
bases. Moreover, when a body of uniform density has an axis of sym- 
metry, we know that the center of gravity must lie on that axis, and 
the problem of finding it is rendered simple. 

When a body having no geometrical center can be separated into 
parts each of which has a geometrical center, the center of gravity 
may be found by the method of moments and "centers of action," 
as already given in Chapter III. If the number of parts is finite, and 
the weight and the center of gravity of each part is known, the center 
of gravity of the whole can readily be found by repeated application 
of the "axiom of moments." 

126. The C. G. of a group. To find the center of gravity of a 
group of bodies whose weights, and the co-ordinates of whose centers 
of gravity are respectively known. 

N. B. In choosing co-ordinate planes, it is generally best to so 
take them that one, and if possible two, of the (C. G.)'s lie in the 
co-ordinate planes. 

1. Let a typical body, whose weight is Wi, have a center of gravity 
at Xu 2/i,2i. Now assuming that the axis of X is horizontal, and letting 
x be the X-co-ordinate of the center of gravity of the group, we 
have by the axiom 



Hence 



Similarly y 



Sn = 



x 2W = Z(Wx) 

H(Wx) WiXj+Wtfk+WiXs+ebc. 
2W Wi+Wi+Wz+etc. 

Wjyi + W 2 y 2 + W s y s + etc. 
W 1 +W 2 + W 3 +etc. 

W 1 z 1 -\-W 2 z 2 + W s z 3 -\-etc. 
Wi+Wi + Ws+etc. 



The point z OJ y oy z , is, of course, the point thru which the resultant 
of the forces supporting the entire system against the earth's attrac- 
tion must act. 

2. When integration is necessary, let dV be a differential element 
of the first order if possible. 



98 



CENTERS OF GRAVITY 



127. To find the center of gravity of a solid cone of revolution. 

1. Take the axis of the cone as the axis of X with the vertex as 
the origin. Fig. 122. Take as an element a lamina* lying between 
two consecutive planes perpendicular to the axis. 
The center of gravity of the element is in the 
axis at a distance x from 0. 
Its volume is dV = 7ry 2 dx; 
Its weight is dW = w7ry 2 dx; 
Its moment about YCY is dM y = w7ry 2 xdx. 




Since 



rx 



y = 



W= — 
h 2 



I 



x%dx = 



Trwr 2 h 



and 



Hence 



M 



ICTTr 2 | ' 

= I x 6 dx = 

h2 X 



7Tivr 2 h 2 



W 4 



2. The center of gravity of a frustum of a cone of revolution is 
readily found, by using the limits hi and h 2i to be 

3 h^-h^ 



4 W-fc 



128. An oblique cone. Suppose we have 
'pyramid, with a base of any figure whose cen- 
troid is known, and we wish to find its center of 
gravity. Let the point C, Fig. 123, be the cen- 
troid of the base, and let VC = x Y . The line VC 
will contain the centers of gravity of all elements 
made by consecutive planes parallel to the base; 
hence the center of gravity required lies on that 
line. Let VX and FF be horizontal axes. Let 
EF be an element parallel and similar to the base, 
so that we have: 

dV= (Area EF) (dxcosO) 



an oblique cone or 




But 



(Area EF) = (Area AB) — , by geometry. 



Xi* 



* An infinitely thin plate. 



THE C. G. OF A SPHERICAL SEGMENT 



99 



Hence W = w — * cosO 

X, 2 

and M y = — 1 (Area AB)w 



C Xl h 

1 I x 2 dx = — 

X 3 



(Area AB)w 



so that 



x = - • Xi 
4 



129. Find the center of gravity of a spherical segment. The 

center of gravity must lie on the axis of 
X if axes are assumed as shown. Fig. 124. 
The element is a thin circular plate. As 
before, we have the general formulas; the 
segment is ABDH. 

dV = Try 2 dx 

dW = 7Twy 2 dx 

dM y = 7rwy 2 xdx 




M 



7 =ttw I (r 2 — 

v = ttw I (r 2 — 
Jk, 



x 2 ) dx = 



7TW 



x 2 ) xdx = 



1TW 



(r 2 -fc! 2 ) 2 -(r 2 -A: 2 2 ) 2 



3 

Xo= - 



(r 2 -h 2 ) 2 -(r 2 -h 2 ) 2 



4 Sr^ki-kO-ikz 3 -^) 
If the segment have but one base, and k 2 = r J 

(r 2 -^ 2 ) 2 



x n = 



4 ^-s^h+h 3 

If the segment be a hemisphere, ki = o 



3 
£ = -r 



It is now obvious that the position of the Center of gravity is in no 

way dependent upon the value of w if it is 
constant. It will therefore for the present 
be considered as unity. 

ISO. Problem. To find the center of 
B gravity of a segment of a thin spherical shell. 

Fig. 125. Let the element be a ring gener- 
ated by the revolution about OX of the 




100 



CENTERS OF GRAVITY 



cross-section rdOdr, or trd9. Its center of gravity is on OX at a dis- 
tance rcosft from YY, and its radius is rsinft Hence 

dV=(trd0) (2?rr shift 



JQ-2 



sin0 dd = QwrH (eosft — cosft) 

= (2?rr) (*) (h-kO^ZTrrht, 
where k 2 — ki = h; 

which agrees with a proposition of elementary geometry, which says, 
that "the area of a segment of spherical surface is equal to the cir- 
cumference of a great circle 27rr, multiplied by the height of the seg- 
ment A." r*Q x 

M y = 27rrH I sm0 cos0 d0 = tt rH (sin 2 ft -sin 2 ft) 

J 02 



tt rH (cos 2 ft — cos 2 ft) 



hence 



a* = - (COSC7i + COSC72) = ; 

2 2 



that is, the center of gravity of a spherical zone is always in the plane 
midway between the bounding planes. 

Corollary. The center of gravity of a thin hemispherical shell is 
at x = r/2, the middle of the central radius. 

131. A segment of a thick spherical shell may be made by two 
parallel planes which cut both spheres; or by two cones which are co- 
axial and have their vertices at the center of 
the sphere. 

The first case is seen in Fig. 126. To find 
the center of gravity of the segment sug- 
gested, find W 2 and M 2 for the segment of 
the larger sphere, as given in 129 above; 
then W\ and M\ for the segment of the smaller 
spheres. The value of x is found by algebra. 



M 2 -M l 

x = 

W 2 -W 1 

which involves nothing new. 

132. In the second case, Fig. 127, when 
the segment is cut from the thick shell by 
two cones, we must integrate an element 
which has two differential factors. The 
shaded areas represent the section of the 





Fig. 127 



SEGMENTS OF THICK SPHERICAL SHELLS 



101 



segment made by an axial plane. The element is a ring whose cross- 
section is (pdd) (dp), and whose circumference is QnpsmO. Hence 
we have, assuming that w = \. 

dV = 2tt p* dp sin 6d6 



W 



Jr 1 J a 



p 2 sin OdpdO 



The order of integration is immaterial since the limits of one variable 
are independent of the other variable. 



Also 



W 



( r 3 ri 3v 

= 2 77 ( I (cos a — cos /3) 

p z sin cos dp d0 

'. 

_ /r 2 4 — n 4 \ /cos 2 a — cos 2 /3\ 



Hence 
Corollary I. 



3 r 2 4 — ri 4 /cosa + cos8 

rp ■ . .1 



4 ri 



3 \ 



If a = and ri = 0, we have the center of gravity of a spherical cone. 



x = -r (l-f-cos/3) 



Corollary II. 

If now B = — y the cone becomes a hemisphere and r— — r as before. 
2 8 

133. Problem. To find the center of gravity of the solid cut 
from the sphere by the cone V-ABC. 
Fig. 128. This problem could easily be 
solved by taking it in two parts : the cone 
VABy and the spherical segment ABC 
(since both volumes and their centers of 
gravity have been found), and then pro- 
ceeding by the algebraic method. It may, 
however, be more simple to proceed directly 
by integration. 

For an element, pass two co-axial and 
co-vertical cones whose half- vertical angles 

are and 6-\-d0. Pass also two consecutive axial planes which make 
with each other an angle d<f>. The solid enclosed will be an oblique 
pyramid whose base is a rectangle (rsin%@d<f)) (rd%0), and whose 
altitude is %rcos 2 6 = h. 




102 



CENTERS OF GRAVITY 



Hence 



dW= -• 4r 3 cos 2 sin 20 dOdxf) 



W = 



S 



ncos 3 



6 sin OdOdcf) 



The limits for </> are and 2tt; for they are and a. 

z 
- I cos s 0sin0 dO 



16nr* T 



^ 



47T>' 3 



(1 — cos 3 a) 



(19) 



The center of gravity of the element is - of the slant height from 

3 4 

V, or -• 2rcos 2 # from YY. Hence 
4 

<Of„ = (-r 3 cos 3 0sin 0d0d<f\ -- rcos 2 



and 



M„ = Snr 



j> 



OsmOdO 



= - 7rr 4 (l— cos 6 a) 



Hence 

Corollary. 

If a= - 






1 — cos"a 
1— cos 4 a 



. r 



7T 




y 

Fig. 129 

about FF is 



the solid becomes the sphere, and as cos — =0 we have 

2 

4 

PF= — 7rr 3 , and x = r which serves as a check 
o 

on the formula. 

136. Problem. To find the center of 

gravity of a wedge cut from a circular 

cylinder. The wedge is shown in plan and 

elevation in Fig. 129. Let the plane of the 

base be the XF-plane, with the center as 

the origin, and the plane XZ a plane of sym- 

2T metry; hence y o = 0. The weight of the wedge 

is known by inspection to be W = — Trr 2 h, if 

2 

w = \. Taking an element (Qyh'dx) its moment 



dM y = Zyxh'dx; 



CIRCULAR AND SEMI-CIRCULAR WEDGES 



103 



but 



hence 



Hence 



h'= — (r+x); and y = (r 2 -x 2 )\ 
2r 



M W -\$y 



W 4 



X- 2 ) 2 {r-\-x)xdx = 



Trhr* 



as was found in Chapter VII for the center of hydrostatic pressure. 

1. When we consider the center of a uniformly varying pressure, 
which is geometrically represented by the wedge-shaped solid ADB 
(Fig. 126); and the center of gravity of the geometric solid, we find 
that the center of gravity of the solid and the center of pressure are 
in the same vertical line. 

2. To find Z , note the fact that a plane passing thru a tangent to 
the base at A, and bisecting the altitude h, contains the centers of 
gravity of all vertical elements, and hence would contain the center 



of gravity of the entire solid. Hence we have, when x= — =x c 

4 

h_ 

4r 



mn 

2 



K)'- 



h 



16 



This problem is closely analagous to that of finding the center of 
action of a uniformly varying force. In fact, some writers picture a 
distributed force whose elements are parallel, as a solid, and find the 
center of action by finding the center of gravity of the ideal solid. 

3. The center of gravity of a truncated cylinder may now be 
found by the algebraic method. 



Problems. 

Ex. 1. Find the center of gravity of a semi-circular cylindrical 
wedge, as shown in Fig. 130. Let y = rcos0, 
x = r sin 0, dx = r cos 9 d6. 

3 



Ans. 



x n — 



Vo 



77?" 



16 

37T 

32 




h. 



Ex. 2. Find the center of gravity of an annular 
wedge which has for a base one half of a ring cut 
from a thick cylinder, and whose upper surface is 




Fig. ISO 



104 



CENTERS OF GRAVITY 



an inclined plane, which passes thru the diameter of the ring as shown 
^ Fig. 131. a _ „. „ 4 




Ans. 



x = 



377 r 2 4 — ri 4 
3tt r 5 4 - 



IL 4 .A 



3_ 



Fig. 131 



^'1"^ 



32 r 2 

137. Composite figures. In the case of a composite 
figure whose components have geometrical forms, the 
center of gravity may be found by the algebraic proc- 
ess of addition. It often happens that certain of the 
components are negative, and that their weight and 
moment must both be treated as negative quantities. 
A single example will now be given to illustrate the 

method when some of the components are negative. 

Figure 132 shows a cube whose edges are 3 feet long. All four upper 

corners are beveled, the bevel being cut away one foot on each of the 

three edges. A central vertical cylinder is cut out 

from top to bottom, the diameter being 1.5 feet. 

The student is to find the volume and center of 

gravity of the solid. 

Suggestions: In finding the volume of one of 

the pyramids cut off, the student should know 

that the perpendicular dropped from the vertex 

of the pyramid upon the equilateral triangular 

base makes equal angles with the edges, and if 

is the angle which the perpendicular makes with 

one of the edges, cos#= Vi. See 06. The student of solid analytic 

geometry may very readily find the length of a perpendicular upon 

a plane which has equal intercepts on three rectangular axes. 

Problems. 

1. Prove that the G. C. of a triangle pyramid is on the right-line 
from the vertex to the centroid of the base, and one-fourth of the 
length of that line from the base. 

2. Find the C. G. of a fine wire having the form of an arch of the 
cycloid generated by a rolling circle whose radius is r. 

3. Find the center of gravity of a solid generated by revolving a 
circle about one of its tangents, thru an angle of 180°. Make use of 
the solution for finding the centroid of a circular arc. 

4. Suppose a hemispherical bowl, radius r, of infinitesimal thinness 
has a plane circular cover, radius r, equally thin, but of twice the 
density (i. e., its w is twice as great). Find the C. G. of the combination. 




Fig. 132 



CHAPTER IX. 

Varying Stress in Beams; Moments of Inertia. 

138. Stress and strain. Modulus of elasticity. 1. All materials 
used in construction are elastic to a greater or less degree. A me- 
tallic bar is stretched by tension, shortened by compression. The 
same is true of wood, stone and cement. 

Think of a steel rod of uniform section, A square inches, and length /, 
subjected to a tension P. Fig. 133. 

When the bar or rod is well made it /^^\\ a" a _J//*-n\ f 
is assumed that the tension at the < — h^-V— 1 C v \Jy~/ > 

cross-section A is uniformly distrib- — * i >^— 

j i • Fig - 133 

uted, so that each square inch sus- 

P P 

tains a ''stress" of— lbs. If we write — =v, we read v as the unit 

A A 

stress. 

2. Under this stress the length I becomes Z + X, if A is the elonga- 
tion of the ro.d, and - is the elongation of a unit of length. The 
elongation a is called the strain, and — is called the unit strain. 

3. Now, so long as p does not exceed a certain limit, called the 
elastic limit, the numerical ratio between p and - is constant. This 

ratio is called "The Modulus of Elasticity" of the steel, and is repre- 
sented by the letter E, so that 



hence 



The last form of the definition shows that if there were no limit 
to the elastic law, a unit stress of E would double the length of 
the rod. 

139. How E is found. The value of E is found by actual tests 
(see 49), in which p has such values as 6,000 lbs., 10,000 lbs. or 
30,000 lbs. A sound bar will sustain such values of p, and when 
relieved, will recover its exact original length. If, however, a stress 

(105) 



^ unit stress 
E — 


= p-r 


I _ 


PL 


unit strain 
p X 
E ~1 ~ € 


X 



106 



VARYING STRESS IN BEAMS 



of 80,000 lbs., or 100,000 lbs. were applied, the bar, when relieved, 
would not fully recover; it would be permanently lengthened, or it 
would have taken a "permanent set." The greatest value of p with- 
out "set" is called the "Elastic limit."* The elastic limit is there- 
fore defined as the stress at which the ratio p/e ceases to be constant. 
With greater values of p, the ratio p/z generally becomes less than E. 
The Elastic Limit, or limiting value of p for sound steel varies from 
30,000 lbs. to 40,000 lbs. The value of E varies from 28,000,000 lbs. 
to 30,000,000 lbs. See Appendix. 

140. Compressive stress. When a bar or block is subjected to 
compressive stress, it is shortened, and if the forces are not excessive, 
the deformation X, will be proportional to p, and we have again a value 

pi 

A 



— = — , therefore E r = 
I E, 



1 [ 



which is the Modulus of Elasticity for Compression. 

Thus there are two Moduli, E t and E c . Under moderate values of 
p, the two E's for steel are assumed equal, tho their elastic limits 
may vary. 

111. Normal stresses. When a cantilever beam is broken by 
an excessive load, its torn fibers in the top of the beam show that 

they were pulled in two by 
a destructive tension; also 
the crushed and bent fibers 
in the bottom of the beam 
show that they suffered from 
a destructive compression. 
These features are easily seen 
if the beam be of wood. A 
steel beam may not break, 
but may bend out of shape; 
still it would show stretching 
on one side and upsetting on 
the other. 

Let us suppose now that 
such a beam, Fig. 134 a, is 
not broken, but is slightly 
bent, and that a thin layer, 
or block, mncf y is distorted 



A 


K 


1 


1 1 




n c 

\Wo 


I 


I s 


1 I 




(a) 




^s 


1 1 




1 




1 1 


m ff 




A 


i - 




1 I 








I 


wT~~~^ 


I I 






\ 





Fig. 134 
(b) 





S 8 


1 


7 f^ r, 


\z 


h 

] 


n 


I 


T- 


u 


w 2 


c 







* A very slight "set" may appear upon the first application and removal of a 
stress below the elastic limit, which set will not be increased by subsequent appli- 
cations tho repeated many times with even greater stress. 



THE MOMENT OF RESISTANCE 107 

or deformed into mnc'f, Fig. b, by the internal stresses. The 
fibers in its top are lengthened, and those at the bottom are com- 
pressed, and the "strain" varies on each side of the center (where it 
must be zero) in such a way that what was a vertical plane face, /c, 
becomes an inclined plane face /V, and the beam actually bends. 
This is what happens under a load, and as the stress is proportional 
to the strain, it is seen that the stress varies uniformly both ways from 
OY, being tension on the upper half, and compression on the lower, 
while the horizontal diameter, across the beam, is a "neutral axis." 
OY is the neutral axis. See cross-section enlarged. 

142. The moment of resisting stress. In the cross-section, 
the rectangle ffcc represents the outer face of the thin layer. All the 
fibers in the area OffY, have been subjected to tensile stress; all in the 
area OYcc have been subjected to compressive stress. This dis- 
tributed action by the portion of the beam, K, has had a tendency to 
turn the thin block round right-handed. In fact, it has turned it a 
little. It is plain that the portion A must resist and balance that 
tendency of K to turn the thin block. 

Let us suppose that the intensity of the stress at a unit's distance 
from the neutral axis OY is a; then the intensity at a point z distant 
must be as; and the amount of stress on an element of the 

s urf ace > dA = bdz, is dF = abzdz. 

And the moment of that stress about the axis Y is 
dM = abz 2 dz 



so 



that M = abfz 2 dz = aj'z 2 dA, 



In the integration the limits of z are — - > and + - ; so that the integral 

2 2 

W A 

becomes — and 

12 njr bh 3 
M = a 

12 
which is known as the moment of resistance. 

When OY is an axis of symmetry the integration could have been 



I 



143. The neurtal axis passes thru the centroid of the section. It 

is readily seen that for every element of the cross-section, dM is 



108 VARYING STRESS IN BEAMS 

positive, since the only factor in it that can be negative (z), is 
squared. It is otherwise with 

dF = abz(dz) 

for z is negative for all points below OY, and it is seen without other 
proof that, if OY is an axis of symmetry of the surface (the cross- 
section of the beam), the algebraic sum of the normal stress, tensile 
and compressive, is zero. In point of fact, F is always zero if the 
beam be horizontal, and is acted upon by vertical forces alone, no 
matter what the shape of the cross-section may be, since the pull and 
the push upon the thin block in the direction of OX must have equal 
magnitudes. In that case, since 

F = abjzdz = 

we must have J zabdz = 

which means that the moment of a uniform stress on the surface 
about the axis OY is zero. Hence, the neutral axis must pass thru the 
centroid of the surface of action. This is a very important fact. 

144. The bending moment. It should be evident from the above 
that the tendency of the external forces, acting on the beam, to turn 
it to the right, is, after some bending, fully resisted by the stresses 
of the fibers within the beam. The load tends to turn the thin block 
about OY one way, while the distributed stress on the rear face, raw, 
tends to turn it the opposite way, and finally to actually hold it. Hence 
the action of the portion A we have called the holding or "Resisting 
Moment." 

Having found an expression for the resisting moment of the beam in 
terms of the internal stresses of the fibers, let us find the resultant 
moment of the external forces which act upon the beam and so call 
into action the balancing moment already found. The external 
forces are W\, the pull of the chain at the end, and W 2 , the weight of 
the beam whose line of action is vertical thru its center. Their mo- 
ment about the same axis, OY, is 

M = W 1 l+-W 2 (1) 

2 

This is called the Bending Moment. 

As the beam, when loaded and slightly bent, is at rest, the two 
moments must balance, that is 

a . b *-=W 1 l+W 2 - (2) 

12 2 



THE BENDING MOMENT 109 

from which a, the internal stress at a unit's distance from the neutral 
axis, can be found as all other quantities are known. 

The intensity of stress is greatest at the extreme fibers. In the case 
of a beam with rectangular cross-section, the greatest stress is 

h 

pi = a. - 

2 

, 2pi 

nence a— — - 

h 

and the equation of moments becomes 

hh 2 7 

Pl — =W 2 - +WJ, (3) 

6 2 

andIF 1 = ^ -El 

61 2 

from which the maximum allowable load, Wi, can be found if p\ is the 
maximum allowable intensity of stress in the fibers of the beam. 

Examples. 

1. Find the extreme fiber stress in a cantilever steel beam, 
£4' long, 2" wide and 12" deep if a 

weight of one ton is hung at its end. ZJZ A _*!'__ 

(Fig. 135.) 

N. B. Reduce dimensions to inches 
and the ton to lbs. Steel weighs about 
490 lbs. per cubic ft. It is evident from 

(1) that the Bending Moment, i. e. 9 the moment of the External forces, 
will be greatest when I is greatest. Hence the cross-section we have 
to consider is at AB, where I is the full length of the cantilever. 

Ans. pi = 17,880 lbs. per square inch. 

2. If pi had been limited to 8,000 lbs. 9 how large could W\ have 
been? Ans. 353i lbs. 

3. A beam 14 /r by 6", and 20' long, supports a uniform load of 

180 lbs. per running foot. 

"||«« Fig. 136. The beam is of wood 
which weighs 38 lbs. per cubic 
foot. What is the maximum 
fiber stress in the beam? 

By symmetry, V 2 = - (weight of beam and load). The greatest 

2 



Fig. 135 



iJ&5&sh*3fed^^ 


A C K 


D F 


M-- 


J , 


Fig. 136 


r* 



110 MOMENT OF INERTIA OF A PLANE SURFACE 

stress is evidently at the center. The external forces acting on the 
half K to make it bend about an axis at C are: F 2 = half of the 
entire load, acting up at B; and half of the entire load, acting down at D; 

bh* lVW\(l l\ , 
hence pi • — = { - — - > etc. 

6 \ 2 / \4 2/ 

145. The moment of inertia of plane surfaces. 
Going back now to 142 

M = ajz 2 (bdz) = ajz 2 dA 

it is evident that while a means intensity of stress at a unit's distance 

from the neutral axis, the integral is only a function of the surface; 

it is the sum of all the products resulting from multiplying every 

elementary area by the square of its distance from an axis in its 

plane. This integral is sometimes called "The Second Moment of the 

Surface," but the more common name is "The Moment of Inertia'* 

of the surface. It is very generally designated by the capital letter I. 

If the axis, from which z is measured, passes thru the centroid of the 

surface, it will be written I . 

Of course / will vary with the size and shape of the surface, for 

example, the cross-section of a beam. In the case of a rectangle, we 

found it to be , , , 

T _ V' 1 

°~n 

We shall soon find values for I for triangles, circles and other 
figures of use in engineering. The general formula which every 
student should keep in mind is 

M = al 

in which M is the Bending Moment of external forces with reference 
to the neutral axis of the cross-section of the beam where the stress 
is to be examined; and al is the Moment of Resistance of the beam 
at the same cross section. 

146. The radius of gyration. As the name "Moment of Inertia" 
was borrowed from another department of Mechanics, which is much 
older in history, so the name "Radius of Gyration" has been 
taken from the same source. It is here defined simply by the formula 

A 



THE RADIUS OF GYRATION 



111 



in which k is the "radius of gyration," and A is the area of the sur- 
face under consideration.* / usually can be separated into two 
factors, one of which measures the area of a cross-section, and the 
other will be, by definition, the value of k 2 . When / is 7 , k becomes k Q . 

In the case of the rectangular beam 



I o = h Jl, A=bh, k 2 = —, k = - V3 
12 12 6 



Engineering hand-books usually contain tables giving values of I , 
A, k 2 for all beams in ordinary use, and pi, the maximum allowable 
stress in the materials used, which greatly facilitates the work of 
designing or calculating the strength of beams. 

147. Take a prismatic beam, whose cross -section is an isosceles 
triangle, suitably loaded with a uniform load, which, with the beam 
itself, weighs w lbs. 



p 


< ^ 


















A 




S X 




v 2 

Fig. 137 
(a) 



"A 




per linear foot. 
Fig. 137. The beam 
is supported in a 
horizontal position 
by piers at the ends. 
Let us find the maxi- 
mum fiber stress at 

a distance x from the end P. The cross-section at A gives the sur- 
face of action, which is shown enlarged. Fig. (b). 

1. Now we know that the greatest stress will be at the top of the 

' . 2 

beam at the point K, which is distant from the line of no stress, — h, 

3 

inasmuch as the neutral axis passes thru the centroid of the triangle. 

Accordingly 



2/i 



Pi 



* The name "second moment" would appear to be more logical than the 
"moment of inertia." The expression zdA is the (so called) moment of an ele- 
mentary area. When this moment is multiplied by z, we seem to get a moment 
of a moment, or a "second moment." 

However, the term "moment of inertia" is in good use and will be retained as 
will the mysterious term "radius of gyration." Both will be accounted for later 
on in this book. All that the students need to know now about either of these 



terms is, what algebraic functions they represent. /= 



Cz 2 dA and k 2 =^- 



112 MOMENTS OF INERTIA OF SURFACES 

So we must find a. Now, we know that 

M = al 

M 
or a= — 

Hence, in order to solve our problem, we must find both M and I . 

2. M is the moment of the external forces acting on the beam- 

Iw 
segment, PA. Those forces consist of the "support" Vi— — acting 

2 

up; and the weight of PA, which is wx acting down, "centered'' at the 
middle point of the segment. Hence, the moment about the neutral 
axis at A is 7 

Tlf wl x w n 2N 

Jsl = — x — wx. — = — (lx — X 1 ) 

2 2 2 

We must next find I , the Moment of Inertia of the triangular section; 
see the enlarged figure in Fig. 137. 

3. For finding I oi take an element (ydz). 

So that we have I = ) z 2 ydz. 

2 
Since 0K= —h, we get by proportion 

o 



^rfiH 



h 2 

The limits for integration for z evidently are and + — h, so that 

o o 

Ih 

( 9. \ 

dz 



•>•*)>-■) 



oh* 



Now, the extreme fiber is at the point K, distant — h from FF, so 
that 

2h M 2h 2 ' \2w n s 

vi= — • — = — • • 36 = (l — x)x 

3 I 3 bh? bh 2 

4. It thus appears that the extreme fiber stress varies as does the 
product of the two segments of the beam: x and l — x. It was shown 
in the calculus that the product of two numbers whose sum was con- 



MOMENTS OF INERTIA OF A TRIANGLE 113 

stant was a maximum when the numbers were equal. Hence, the 
maximum value of p occurs when x = — which gives 

, % , , Swl 2 

(Max.) Vl = -— 

bh 2 

148. Engineer's hand book. Of course, we knew, without 
mathemathics, that the greatest stress (greatest danger of breaking) 
was at the middle of the beam, and had we turned to the Engineer's 
hand book we should have found 

wl 2 bh? 

Max. Bending Moment = — \ I = — > 

8 36 

1 k 2 
A=-bh; h 2 =—\ 

2 18 



and if we put c= —h, and M = — • Ak 2 = al t 



0> 



3 
we should have had, as found above: 

wl 2 3 bh h 2 . Swl 2 

— =Vi' — - • — - so that pi = 

8 ah 2 18 bh 2 

149. Some relative properties of the areas and stresses above 
and below the neutral axis of a triangle.* The reader has noted the 
fact that the algebraic sum of the normal stress on a cross-section of a 
horizontal beam is zero. This is self-evident if the axis YY (the 
neutral axis) is an axis of symmetry; it does not look quite so evident 
if the section is a triangle as in Figure 137. However, if we find the 
total stress (tension) above the neutral axis; and then the total com- 
pressive stress below the neutral axis, and find them of equal magni- 
tude and with opposite signs, the proof of the doctrine laid down 
long ago (that the algebraic sum of the horizontal forces must be zero, 
because the beam is at rest) will be complete. 

1. Taking the general formula for the stress, Fig. 137 (6) 



F = a\ yzdz 
Substituting y= — ( — h — z\ 



* While the discussion which follows may not seem to be very practical, it deals 
with general principles which are vital, and in an emergency may be important. 
Engineers have been known to quarrel over matters here discussed. 



114 



MOMENTS OF INERTIA 



and integrating for the surface above YY we have 



*-tfG 



h-z) 



zdz 



\ 3 3/ 



a b I hz 2 
Ti 



4aW 
81 



(1) 



2. If now we integrate for the stress below YY, we shall have 

6 (1 
9 'h 



C-H 



as before. 



(It must be remembered that z is itself negative.) 
Accordingly we have 



h J_A^ 3 ' h ^ 3 3 ' 



4aM 2 
81 



and jPi + F 2 = 0, as was expected. 



The proposition is universally true if the neutral axis passes thru the 
centroid of the surface. 

3. But the Moments of the stresses above and below are not equal 

for the triangle, as is readily shown. 

Taking the formula r 

M = aj yz 2 dz 

and integrating for the moment of the stress above YY, we have 



Jfi=— (-h-z\&dz=— abh* 
h J o \S J 243 



4. Integrating for the moment of the stress below the neutral axis 
we have 



if* /«, \ 
= a— I — h — z 



4 ) 

1/ 



11 

972 



abW 



so that — - 1 = — * Mi being the larger. 
M 2 11 

r -ir> 

Since a is the same for the both areas, — = — 

h 11 



(23) 



THE STUDY OF A TRIANGULAR CROSS-SECTION 

bW 

bh? 



115 



The sum M x + M 2 = M = a 



36 
/i+/ 2 = / = 



36 



and 

as already found. 

5. The "Center" of the uniformly varying stress above the neutral 
axis is 



2l = 



~abh 3 
Mi 243 h 



F 1 



— abh 2 
81 



which is just half-way from YY to K. 

6. The "Center" of the uniformly varying stress below the neutral 
axis is 



22 



M2 

F 2 



11 
972 



ab¥ 



abh 2 



11 

48 



so that, numerically, 



81 



9 



3 ^ 48 16 



It must now be evident that the internal normal stress of a beam 
forms a couple at every section whose force is Fu whose lever arm is 
L, and whose moment is 

^ T 4<ahb 2 vv 9 7 abh s ,, T 
FiZ= X — h= =M = aI Q 

. , , , 81 16 36 

as already lound. 

150. Results may be summarized. 

1. Ai is to A 2 as 4 is to 5. 

3. i/i is to lf 2 as 16 to 11. 

4. 7i is to I 2 as 16 to 11. 

5. 0d= —center of stress on Ai. 

3 

11 



6. 0C 2 = — ^> center of stress on A 2 . 

48 

7. OCi + OC 2 = X 



I 

/ c 
F / 


\ Y 


/ A F * 

/ A 2 


>C 2 M 2 \ 


( 


Fig. 138 



_9_ 

16 



8. LF = M 1 + M 2 =M=a 



bh 3 
36 



116 MOMENTS OF INERTIA OF SURFACES 

A similar set of results and comparisons can, of course, be made for 
every cross-section of a beam whenever the areas on the two sides 
of the neutral axis are unequal, which is often the case, because the 
material. used is not as strong and elastic in tension as it is in com- 
pression, as, for instance, when the material on one side of the neutral 
axis is concrete and on the other side of the neutral axis is steel, or 
a combination of steel and concrete. Such cross-sections will be 
considered in later sections. 

151. This chapter does not aim to cover the extensive subject 
of beams. Some simple forms have been introduced to illustrate the 
subject of uniformly varying stress, and to develop the function, 
"Moment of Inertia," and show how intimately connected it is with 
the study of stress and the strength of beams. 

1. Before solving and proposing a variety of problems for finding 
values of /, it may be well to state again some assumptions we are 
making, and to put down in words and in the language of mathe- 
matics, the general equations we are to apply. 

2. All beams and girders are assumed to be of homogeneous material, 
whatever may be the cross-section, and to have uniformly varying 
normal stress at every cross-section, when they are slightly bent under 
moderate loads. In all cases this important formula holds: 

The Bending Moment of external forces acting to bend the beam 
at any section is numerically equal to the Moment of Resistance of 
the internal stresses acting at that section. 

If p is the stress in the extreme fiber, c the distance of the extreme 
fiber from the neutral axis, and M the Bending Moment, the above 
dictum is more accurately expressed as follows: 

V t vA 
M = aI = F -I =^- . W 
c c 

3. The general formula for the Moment of Inertia when YY is 
the axis of reference is 



= J J z 2 dzdy. 



When, however, the element of the surface can be a differential strip 
parallel to the F-axis, the formula is greatly simplified by assuming 
the element ydz, so that we have 

I = jz 2 ydz. 

152. Moments of Inertia, fundamental examples. 



MOMENT OF INERTIA OF CIRCLE AND ELLIPSE 



117 



1. Find I for a circle whose radius is r. Fig. 139. The element 
of the surface is ydz s* r 

7 = 2 z 2 ydz 

Introducing 0, we have ° 

y = 2r cos 9. z = r sin 0. dz = r cos odu 



and I n = 2r 4 I sin 2 cos 2 9d9 



f 

= - f 




Fig. 139 



sin 



%9d{%9) 



TTr 



1= — =A- =Ak* sothat&„ 2 =- 



4 4 

For a semi-circle about its base 



(25) 



7rr 4 
~8~ 



2. The I of a Circular Ring, radii r\ and r 2 . 
7 



7rr 2 4 7rr! 4 7T, r 2 2 +ri 2 
= -(r 2 4 -ri 4 ) =A. 



4 4 4 

3. Find I for an ellipse. 

'II z 

Let the equation of the ellipse be - + - =1 and let the axis of ref- 

b 2 c 2 



erence be YY. 




From Fig. 140, it is evident that 

c 

!„ = % \ z 2 ydz 



'•-i 



Fig. 140 



Introducing 9, we have 

y = b cos 9. z = csin9. dz = c cos Odu 
so that 



f sil 



I o = 4<c 3 b \ sin 2 9 cos 2 OdO 

_7TC*b _ c 2 _ '. , c 



(26) 



118 



MOMENTS OF INERTIA OF SURFACES 



Y 


V 


Y 






Z \ 

y— \ 


h 




/ f dz 








7.1 
1 






< 



4. The Moment of Inertia of a triangle about an 
axis thru its vertex parallel to the base. Fig. 141. 

The element is ydz, in which 



Fig. 141 



Y= - 
h 



hence 



If 



>dz = 



blv 



= A. 



Ak\ :.]<= JL, 



h 

V2 



(28) 



(29) 





z I 




z 


i .Hi 


z 


z 


1 1 


z 



b=i2 




5. Had we used the base as the axis of reference, we should have had 
hJ o h\3 4/ 12 

6 

7. The relations of the three values of 
/ for a triangle are thus shown in Fig. 142. 

153. An exterior axis in the plane. 

The Moment of 
Inertia of a sur- 
face with refer- 
ence to an axis 
not passing thru 

the centroid may be found directly by inte- 
gration, as has just been shown, but it is 
often more convenient to make / depend 
upon I for a parallel axis. Thus : 
To find J for the axis ZZ, Fig. 143: 

(so that dy = dy>). W^> but 2/ = c +^ 

If Z Z is parallel to ZZ, and the distance from ZZ to the centroid is c, 

HenCe I z = f(c*+2cy'+y'*)zdy' 

= c 2 fzdy'+2cfy f (zdy') +jy' 2 zdy' 
= c*A+I 

Since the first integral is plainly the area of the given surface; the 
second integral is the moment of a surface about its own centroid 
axis, which must be zero; and the last is merely I . Hence the rule: 

I z = I„+Ac* (1) 



Fig. 143 



MOMENT OF INERTIA FOR A PARALLEL AXIS 



119 



The Moment of Inertia of a surface with reference to any axis in its 
plane is equal to I for a parallel axis thru the centroid of the surface, 
plus the product of the given area multiplied by the square of the dis- 
tance between the two axes. 

1. Thus the I v for the triangle in 153 could have been found from 
the knowledge of 



° 36 



for / v= ^ + ^.(^> = 6/,s(_L + A) = 

36 2 ' \3 / V36 36/ 



bW 

T 



2. Conversely, knowing I z , I for a parallel axis can be found from 



the formula 



J =L-Ac* 



bh 3 
3. For a rectangle about its base we have, since I = — 

12 



12 



Wi = ft i! 



4. For a semicircle about Z Z 
Fig. 144, z 



Hence 



/4r \ 2 



vrr 4 7rr 2 16r 2 r 4 /rt „ _.. A I 

J = — = (97T 2 — 64) =^4. 

8 2 9?7 2 72tt V 



3677 



(2) 





No. 2 



Fig. 145 



of the formula: 



5. Finally, knowing / and c for any axis, 
we can find I for any known parallel co-planar 
axis. Suppose we have I\ and c for No. 1; 
and we wish to find Ii for No. 2. Fig. 145. 

I = h-A c * 

154. The Moment of Inertia of a compos- 
ite surface. It is first necessary to find the 
centroid of the surface, then I is found by use 

I for Z Z = XI +X(Ac>) 



The addition in practice is algebraic, since both I and A may be 
negative, as will now be illustrated. See Fig. 146. 



120 



MOMENTS OF INERTIA OF SURFACES 




Fig. 146 



Example. The line YY is an axis of 
symmetry, and must contain the cen- 
troid G . Take any convenient line as 
z a preliminary Z-axis as Z'Z' . 

The total area is 

2A = — +&!&!- 2&A+ - Ihh 

2 2 

The total moment with reference to 
Z'Z' is 



4r\ , b\h\' 



, a \ 7Tr ( i , 4r 1 , ^"1 U h 2 



6 



In the case of the two absent rectangles, both areas and moments are 
negative. In the case of the triangle the area is positive but the 
moment is negative. Hence 

nr 2 (11,+ —) + W-2W- ^ 
X(AY C ) _ Sir I 3 

HA 



Vo = 



Trr 2 + 26i/ii — 46 2 /*2 + &i^3 



Having found and located G , find the I for the areas separately, 
with reference to their own centroid axes. Then find HAc 2 and add 
results. 
Numerical example — 

Let r = 2, b x = 6, h = 4, h 2 = 3, h 3 = 3. 

Find the position of G and the value of 1 . 

155. Built up beams, struts and columns or posts are very fre- 
quently made of separate steel bars or plates securely riveted together 
so that the cross-section consists of parts, and the moment of inertia 
must be found as in the last example. 

When a post or column is used to carry a vertical load, its stiffness, 
whereby it resists lateral bending, is an important element in its strength; 
and its moment of inertia is needed with reference to more than one 
axis thru the centroid of a section. For example, in the case of an 
ellipse (Fig. 140) we have 



r y = fz 2 dA, and I z fy 2 dA 



N. B. — When two Moments of Inertia are found for the same 
surface, about different rectangular axes, with at the centroid, 



COMMERCIAL SHAPES OF STEEL 



121 



the larger will be called J or I l9 and the smaller may be called J or 7 2 . 
Thus for the ellipse whose major axis is 2a, and whose minor axis 
is 26, , ro 

T 77GTO , T TTab 6 

ii= and J = • 

4 4 

156. Commercial shapes of rolled steel. In determining shapes 
of rolled structural steel, exact dimensions are given including rounded 
corners and fillets, as 
is shown in Fig. 147; 
but in computing 
Moments of Inertia 
for strength and 
weight they are often 
ignored or their fillets 
are assumed to 
balance the rounded corners, as in Figs. 148-9, which are taken with 
notation and formulas from the "Carnegie Pocket 

?-*■*> Book." 



^W"' 



Fig. 147 



6f5" 



<- 



-b ^ 

Fig. 148 



Let the student check the values 
of A, I and J. 

7 = Mom. In. for a neutral axis 
s3;t____] parallel to a flange. 

J = Mom. In. for a neutral axis 
parallel to a loeb. 

^4n7-Beam. Fig. 148. 
ATe& = A=dt+(s+y)2z 



Fig. 149 



1 


2 


• 






1 


6 3 (rf-/0+/* 3 + — (6 4 -/ 4 ) 




A Channel Section. Fig. 149. 


Area = dt+(s-{-y)z 


11 
X ^A 


b2s +l ht2 +~(b-ty(b+2t) 


12 


bd?- -(h*-l 4 ) 

4 




3 


2s6 3 + Z* 3 + — (6 4 - 
12 


-*) 


-A 


v 2 



122 



BEAMS AND COLUMNS IN ACTUAL USE 



•Q 



T 



"I 



?pk /#* 



—L. 



Fig. 150 




tit 



Fig. 152 



Fig. 151 consists of: a 
Web-Plate 60"x%"; each 
Flange has 2 Angles 
6"x6"x%", and 3 Plates 

14"x%". 

Fig. 153 is a cross-section 
of a Column of the Chicago 
& N. W. R'y Office Build- 
ing, Chicago; 6 Plates 
16"x%"; 2 Plates 16"xiVi 6 "; 
2 Plates 12y 2 "x%"; 4 Angles 
6"x6"x%". 




Fig. 151 



Fig. 154 



Fig. 155 



Fig. 154 is a cross-section of Column 280, Waldorf-Astoria Hotel, New York. 
It consists of: 10 Plates 32 1 /2 ,/ x% ,/ ; 4 Plates 36"x%"; 4 Angles 6"x4"xHi 6 "; 
8 Angles 6"x3y 2 "x%". 

Fig. 155 is a cross-section of a Column in Chicago Steel Co. Building, Chicago. 
It consists of: 2 Plates 32 ,/ x 1 / 2 "; 2 Plates 23 ,, x%"; 4 Plates 18"x% 6 "; 2 Plates 
16"x% 6 "; 4 Angles 6"x6"x%"; 12 Angles 6"x6"x% 6 "; 4 Angles 6"x4" x y2". 



LOWER CHORDS OF THE QUEBEC BRIDGE 



123 



Fig. 156, is a cross-section of 
the lower chord in the Quebec 
Bridge which failed during the 
erection. 

It consisted of materials as fol- 
lows: 

Outer Ribs: 2 Plates 54"xi% 6 " 
Each 1 Plate 54"x% / ' 

1 Plate 37%"x 15 /i 6 " 

2 Angles SWx 1 ^" 

Inner Ribs: 3 Plates 54"x 13 /i 6 " 
Each 2 Plates 46 // x 15 /i 6 ,/ 

2 Angles ^'xdV^xWie" 

Lattices across the top and 
bottom of all four ribs. 



H 




L.JL 



Fig. 156 



10 -3" Overall 




Double Lacing 8 



Fig. 157 



Fig. 157 is a cross-section of the lower chord of the new Quebec Bridge, as 

designed for the position of the one that failed. The exquisitely drawn cut, is 
almost self-explanatory. However, the section requires : — 



8 Plates 81.5"x%" 

8Plates81.5 /, xl 1 / 8 // 
16 Plates 20"xl" for flanges 
16 Angles S'^'xl" for ribs 
10 Angles 6"x&''x%" for diaphragm plates 

2 Plates 33 /, x 11 /i6 // for diaphragms 



4 Plates 33"x%" vertical between ribs 
16 Angles 6 // x4 // x 1 / 2 // for vertical plates 
8 Angles 4"x4"x 1 / / 2 // for diaphragms 
4 Lattice Lacings 8"xr' 
3 Plates V2" thick horizontal between 
central ribs 



124 



MOMENTS OF INERTIA 



157. A variety of examples showing cross-sections of beams and 
struts in actual use is given above. The moment of inertia of each one 
may be found from the dimensions given. 

Remarks upon Figs. 156 and 157. 

The chord member which failed, A9L, had lattice bracing along the 
edges of the ribs as shown by the broken lines. The purpose of this 
bracing was to prevent the buckling of the ribs sideways. Prof. 
Geo. F. Swain, of Havard, who made a careful study of the design, 
is of the opinion that failure was primarily due to the inadequacy of 
that bracing. 



CHAPTER X. 



Moments of Inertia — (Continued). 

The Relation Between Strength and Moment of 
Inertia. 

158. The strength of a beam is measured by the load it can 
carry with safety. The test of strength may come from a load at the 
end of a cantilever, or at the center of a beam supported on piers. 
In either case the bending moment is proportional to the load; hence 
M, the bending moment, or its equal, the Moment of Resistance, is a 
measure of its strength. It was found that 



M 



'Pi 



in which c is the distance from the neutral axis to the extreme fiber, 
and p\ is the greatest intensity allowed. It follows that the strength 

of a beam is directly proportional to I, 
and inversely proportional to c, p\ being 
the same. 

1. To illustrate, suppose a beam 
has an elliptical cross-section and that 
it is loaded first with its major axis 
vertical, and second with that axis 
horizontal. Fig. 158. 




Fig. 158 



STRENGTH EDGEWAYS VS. FLATWAYS 



125 



We have 



M l = — • =p 1 (7rab) • - 

a 4 4 



' pi iraib* b 

M 2 = — • =p 1 (7rab)- 

b 4 4 



Hence 



Ml 
if 2 



or the strengths are to each other as the semi-axes. 



2. Again take a rectangular beam. 
Fig. 159. When it is "edgeways" 



Ml =^.™ 



Pi(hb) - 
12 6 



i 



Fig. 159 



When "flatways" 

,, %vi hb 3 ,n\b 

M 2 = -f 1 = p 1 (hb)- 

b 12 6 

so that 

Jf i _ h 

W 2 ~ b 

This means that a floor joist 2"xl2" will safely support six times 
as much when on edge as when flat. This is abundantly confirmed 
by personal experience. Proper lateral support is assumed. 

3. A third case is hardly a matter of 
experience. Suppose a square beam is placed 
J with sides vertical. Fig. 160. Its 
strength is 



Tig. 160 




s 



S* 



Pi(s 2 ) 



If its diagonal is vertical, 



M 2 =^ -2. < ^-=p 1 {2c*)- 
c 12 6 

Mi = * 5 
M 2 c 

or the strength of No. 1 is 1.4 times that of No. 2. 

The unmathematical reason is plain: In No. 1 the whole extreme 
layers of fiber are utilized for the stress pi; in No. 2, only a single fiber 



126 



THE STRENGTH OF AN 7-BEAM 





4- 


-4'- 


> 






1" 






i 




< 4-- -3 










1" 


1 






(.-!-% 




i/' 




















* 



Fig. 161 



on the top and the bottom has the stress pi while the 
increased material near the horizontal diagonal adds 
very little to the strength. 

156. The moral derived from the above illustra- 
tions is, to put as much of the material as far from the 
neutral axis as possible, leaving near the neutral axis 
only enough to preserve the unity and stability of the 
beam. This will be illustrated by two ideal examples. 

1. Compare the strength of a steel bar 4" square 
with an /-beam made of the same amount of steel, the 
flanges being l"x4", and the web l"x8". Fig. 161. 



3/i = p : 



bh' 



M, 



Pi 



. 1 . / 4 ( 10 1 3 _ 3.(8)* 



5 V 12 



12 



= Pl 10-for No. 1. 
3 



) = pi41 — for No. 2. 
/ 15 



Hence 



M2 



41 



15 



4 nearly. 



10 



or, the I-beam with the same weight of material is almost four times 
as strong. 

2. A certain load, W, is to be carried by a cantilever beam of 
length /. Compare the weight of a beam in the shape of a solid cylin- 
drical rod, with the weight of a cylindrical tube of the same length, 
with an exterior diameter n times as great, and equally strong. 

The expression for the strength of a solid cylinder is 

M c = pi(iTr 2 )- =piAi- 
4 4 

If the exterior radius of the tube is nr, and the inner radius #, we 
have, 



M t =^--{I nr -I x ) 



nr 



(n 4 r A — x 4 ) 



Equating the two expressions for M , since they are equally strong, 

and solving for f,we have 

x 2 = r 2 vn i — n 

Hence, the area of the ring, which is the cross-section of the tube, is 

A 2 = 7r(n 2 r 2 -r 2 yln*-n) 



COMPARE THE TUBE WITH THE ROD 



127 



Since the weights of two beams of the same length are to each other 
as the areas of their cross-sections, we have. 



Weight of solid cylinder 
Weight of tube 



A 2 



irr 



7rr 2 (n 2 
Zn 
1 



— V?! 4 — n) " 
nearly 



(since 



Vn 4 — n 



.2_ 



2n 8n 4 



etc.), which means that 



If the tube's exterior diameter is 



( twice 



I as great as that of the 

solid cylinder, and equally strong, it will weigh only one ] ? Ur , [ 

as much; and so on: under ideal conditions, which would be quite 
unreal if n be made too large. 

The saving of weight while preserving strength, thru the use of tubes, 
is the secret of light-framing and light-rolling stock. For a discus- 
sion of the stiffness of hollow shafting; see Chapter XIX. 

157. The axis for the maximum value of 7 . The student is now 
prepared to see that while an unsymmetrical surface may have an 
indefinite number of values of I for different axes across it, there always 
must be one greatest 7 , and one least. In engineering practice both 
the greatest and the least are very important. It is now necessary to 
show how they are found and valued. 

Let 0, Fig. 162, be the centroid of a plane 
surface, and OZ and Y a set of rectangular 
axes so taken that I z and I y can be, and 
have been, found. Let the larger be I z . Let 
0Z\ and 0Y\ be a second set of rectangular 
axes meeting at 0, and determined by the 
angle 0. Let P mark the position of an element 
of the surface, d 2 A, whose co-ordinates are 
(z,y) and (zi, yi) for the respective axes. Fi s- 162 

The quantities I z = j I y 2 d 2 A and I y = ( I z 2 d 2 A are supposed to 

have been found. The surface element d 2 A may be thought of as dydz, 
or dy'dz'. 

From the figure, 

yi = PQ — RN = ycosO —zsinO 
Zi = RQ-\- ON = y sin 0-{-zcos0 




128 MOMENTS OF INERTIA 

Hence I el = j J yW 2 A = J J (y 2 cos 2 0-2yz0sincos0+z 2 sin 2 0)d 2 A 
I zi = cos 2 0l z — 2sin6cos0\ I yzd 2 A + sin 2 9 1 y . 

Similarly I yl = sin 2 I z + 2 sin cos $f j ijzd 2 A + cos 2 I y . 

The integral J J yzd 2 A is sometimes called (for the sake of a name) 
the Product of Inertia. More conveniently it will be called simply K 

K=ffzyd?A. 

When necessary its value will be found by integration. If either 
OZ or Oy is an axis of symmetry, K is zero, since the differential terms 
in the integral are in pairs which balance. 

Therefore, we have 

I zl = cos 2 0I z -2s\n0cos0K + sin 2 0I y (1) 

^^si^^^ + gsm^cos^Z + cos 2 ^/^ (2) 

Adding, we have 

l zx +I yi = I z + I y (3) 

Knowing I z and l yy and having calculated 7 a i, I y \ can be found from 
the last equation. 

1. Take for example the simple case of an ellipse and suppose we 
want l z \. 

We know that T irba 3 , T irab 3 

i = ' and i 7 . = 

4 v 4 

and by inspection we see that K = 0. 

Let 0=^ 

4 

Then I zl = **±** = I yl = irab ■ ^±^ 



Interesting values follow if 0= -and if 0= — • 

6 3 



MAXIMUM AND MINIMUM VALUES OF I 129 

Going back to Eq. 1, we see that the value of I zl varies as 9 varies, 

and will have either a maximum or a minimum value when — ~ is 

do 

zero. Hence, differentiating (1) and remembering that I z , I y and K 
are constants, we have 

^ = -2s\n9cos9l z + 2(sitf9-cos 2 9)K + 2s\n9Gos9l y 
du 

or ^% = ^sintfcos^/.-Z^+gtsin^-cos 2 ^ 

du 

Placing the second member equal to zero so as to find the values of 

9 for the greatest and least values of I z > we have 

{I z - I y ) sin 29= -2 cos 29. K 

tan 20=-—^ (4) 

If K = 0: 20 = and it; l z~ 1 v 

or 9 = and — » hence in the case of the ellipse, the major and 

2 

minor axes of the ellipse are the axes for minimum and maximum values 
of 7. 

158. When K is not zero. The angle 20 always has two 

values differing by 77; hence 9 has two values differing by -, one acute, 

2 

the other obtuse, so that the axes for greatest and least values of I are 
at right angles. In all cases, since I z >I yi the axis for maximum 7, 
makes an angle numerically less than 45° with the original Z-axis. 
The axes for /(max.) =7; and 7 (min.)=J, are called the Principal 
Axes of the Surface, 

Since K = for an axis of symmetry, it follows that that axis and 
its perpendicular thru the centroid are the Principal Axes. 

159. The value of Kt for the axes OZi and 0Y U 
Substituting the values of yi and Z\ from (157) in the expression 

K\= I I yiZ\d 2 A, we have 

K 1= f JV sin0 cos9 + 2y (cos 2 -sin 2 0) -z 2 sin# cos9]d 2 A 

K, = sin 9 cos 9 (I, - I y ) + (cos 2 - sin 2 9) K 

Comparing the value of K x with the value of — -^ we see that 

so that, when dl zi is zero, Ki = and conversely. 



130 



MOMENTS OF INERTIA 



Hence, the "Product of Inertia* ' is 
zero, for the Principal Axes, without 
regard to symmetry. 

159. Angle irons. It is frequently 
necessary to know the maximum and 
minimum values of I for a cross-section 
of an "Angle Iron" shown approximately* 
in Fig. 163. It has no axis of sym- 
metry, and the solution of an illustra- 
tive problem will be best with numer- 
ical values. 

The centroid is found by taking 
moments of areas about AB and AC. 

The centroid is distant from AB 1.42", and from AC, 0.83" 

The position of the centroid is shown at o. 

160. How to find I z , I y and 

K. They can, of course, be found 
directly from the figure by means of 
the formula for parallel axes, but z ~ 
the following is more simple.** 

1. Draw two axes, Z' and Z", 
parallel to OZ, and thru the centroids z _ 
of A\ and A 2 respectively. Fig. 164. , 




Fig. 163 



Then I z = I zl + A 1 s 1 2 +I z » + A 2 s 2 2 



J A S 



But 

hence 



A\S\ = AzS2, ands = Si + s 2 , 

A 2 s , , Ais 



Si = 



Ai+Ai 



ands 2 



A±+A 2 



Fig. 164 



Y' 



so that 
and 



Ais 1 2 +A 2 s 2 2 = 



A 1 A 2 2 s 2 +A 2 A 1 2 s 2 A 1 A 2 s* 



I s =Iz>+Iz.+ 



Ui+i 2 ) 2 

AU 2 s 2 



A,+A 2 



(1) 



A x +A 2 
In the same way it is readily shown after drawing the axes Y' and 

A 1 A 2 r 2 



F" that 



J-y J- y> I" J- y" T" 



Ai+Az 



(2) 



* In angle irons as actually used, the salient angles B' and O are rounded, and 
the re-entrant angle D has a fillet. See Fig. 147. 

** This elegant analysis appears to be due to Muller-Breslau. 



THE MOMENT OF INERTIA OF AN ANGLE IRON 131 

2. To find K= f f yzd 2 A, we take the parts A\ and A 2 separately 
so that K = K' + K". We see that 
y = y'-si 

% = %' — T\ 

and yz = y'z' — r 1 y' — s 1 z'-{-s 1 r 1 

hence K' = ffy'z'd 2 A - rJfy'd 2 A - Sl f fz'd 2 A +*n f fd 2 A 

But the axis Z'Z' is an axis of symmetry of A u hence 

§fy'z'd 2 A = 

As the axes z' and y' pass thru the centroid of Ai, we have 
j"jVdM = 5 and // z'd 2 A = 0; and sinceJ*J"d 2 ^i = ^ 1 
we have IsT^SiTvli and similarly K" = s 2 r 2 A 2 

But, as before si = , and s 2 = 



Ai+A 2 A±+A2 

A 2 r , ^4ir 

ri = , and r 2 



At+A 2 Ai+A 2 

hence K = K'+K"= A ^ Sr + ^ lM ^ 



(^! + ^ 2 ) 2 Ul + ^ 2 ) 



and Z = ^ (3) 

3. The numerical values of I Zi I y and K are now easily found. 
om Fig. 163 we 1 
and from figure 164 



5 

From Fig. 163 we have ^4i = 2, A 2 = - 

4 



Hence 



5=2—— = - and r = — -f- — = - 
V 4 / 4 4 4 2 

1(4)3+ Wl) 3 2X 5 -X^ 

,2 2 V2/ 4 16 

i 2 — -|- ■ . — . 

4 
7 z = 2.696 + 2.356 = 5.052 

T - ^ 2 V2/ 4 4 

4 



132 MOMENTS OF INERTIA 

7^ = 0.692 + 1.731 =2.423 

sx 5 - x «I 

4 8 
K= =2.02 

2+* 
4 

4. It will be noted that all these quantities, tho they refer to axes 
meeting in the centroid of the surface, do not involve a knowledge of 
where that centroid is. The quantities s and r are the co-ordinates of 
the centroid of A^ with reference to the axis meeting at the centroid 
of A\. When, however, we calculate the angle for a principal axis, 
and proceed to draw that axis, the centroid of the entire surface must 
be used as found above. 

The angle 6 is determined by formula 

tana^^gg- - 404 — 1JM 

I g -Iy 5.06-2.42 
%0= -5.6° 50' 

#=-28° 25' 

Having the numerical values of I z , I y , K, 9 and the co-ordinates of 
the centroid of the entire surface, we can plot the center, draw the 
principal axes and compute the values of I (max.) and (J min.). 

5. The computation is somewhat simplified by combining equa- 
tions (1) and (2) 157 , with the equation 

tan 20=—^ 

and we get z v 

I = I z -Ktsin0* (4) 

Combining this last equation with Z+ J = I z -\-I y , we get 

J = I v + Kt3Lii0 (5) 



* Eq. 4, 157, gives 

Iy — Iz + tSLn2 z_t ~ 2sin0cos0 

Substituting in (1) we get 

/„ . sin e (cos 2 — sin 2 \ 

/2 sin cos 2 — sin cos 2 + sin 3 0\ r 
=I Z - K { ~— ^ =I Z -Ktane 



THE VALUES OF I AND J 



133 



Since i£ = 2.02 and tan (-: 
7 = 6.145 
J = 1.33 
13 



since Ai-\-A 2 = 



and k 2 = 



25') = —0.541, we have 
fci 2 = 1.88 
A; 2 2 = 0.41 

k being the radius of gyration. 



4 Ax+A 2 

101. When .the angle-iron which we have been studying is used 
as a beam, the axis for I should be horizontal, and the distance to the 
extreme fiber should be computed or measured from an accurate draw- 
ing. See Fig. 163. Then the greatest Moment of Resistance the 
bar is safely capable of is 



M, 






Axh 2 = 



Pi 



(1) 



However, if an angle-iron is used as a strut of post, it is liable to 
bend by buckling, and as the buckling is always in the plane of least 
resistance, it is only the least Moment of Resistance which is to 
be counted upon. 



M ■ = £ 

■"-*■ mm 



Pi 



(2) 



C 2 C 2 

Steel "angles" may be rolled of various dimensions, with some 
allowances for round corners, and a fillet, but, in fact, only certain 
standard sizes are in the market. 
A general formula for hi and k 2 is 
hardly worth while here. 

162. When the arms of the 
angle iron are equal, as in Fig. 16.5, 
there is an axis of symmetry, AD. 
Consequently, K = 0, and AD and 
the perpendicular thru the centroid 
of the surface are the principal 
axes. The Moment of Inertia, I z ,, 
is readily found by subtracting the 
moment of inertia of the triangle 
A'B'D, with reference to the 

axis OZ, from the moment of inertia of the triangle ABD, and doubling 
the result. The moment of inertia, I y9 is found by subtracting the 
moment of inertia of the small square from the moment of inertia 
of the large square, after finding the centroid of the surface. 

The distance OC is found from the equation 

(0C)[(6) 2 - (5) 2 ] = (5) 2 Vi = (5Y(CC') 

Using the dimensions shown in the figure, find I z and I y . 




Fig. 165 



134 



MOMENTS OF INERTIA 



163. A Z-bar. A Z-bar usually has equal flanges, and tho the 
cross-section has no axis of symmetry, its centroid is readily seen. 

With the axis and dimensions as shown 
p in the figure (Fig. 16G), find the Principal 
Axes and the max. and mih. values of 
M = al. 




L = 



cb 



+ 2 



0& 



12 



+ acX 



C-rJ) 



I z = +-ac(6-c) 2 

12 2 

bc?+%a?c . ac 

/„— - H (a + c) 2 



12 



2 



P' Fig-166 



K= -ac(b-c)(a + c). See (3) 160. 



Hence 



fc z 2=li, and V= If 



Now, as 



tan 20 = 



2X 



the angle #, which is a negative angle, is readily found, as are the 
principal moments of inertia: 

/! = /,-# tan 

J 2 = I y +K tan ft 

11 3 

Let a= — ' 6 = 6, c= -> inches, so that A=Ai-\-A 2 +A 3 = 8.Q^5 

4 4 



Then / 2 =-L( 3 -x 2 16 +2 xHx^ + i(iix^X^)= 42 .l* 
12 \4 4 64/ 2 V 4 4 16 / 

7s= J_( 6X ?I +2X l^X 3 -) + ixiix 3 -X "-1M4 
y 12 \ 64 64 4 / 2 4 4 16 



X= I X lix 3 X?i-X 1=18.95. 

2 4 4 4 2 



* K= AiA 2 sr ^ Ai=A 2 = ac, s = b — c, r = a + c. 



A1+A2 



THE MOMENT OF INERTIA OF A Z-BAR 135 

42 12 

Hence k z 2 = — ^— =5.00, so that fc 2 = 2.24 

8.625 

1 5 44 
and K=— — = 1.79, and fc= 1.34 

y 8.625 v 

tan 20 = -1.42 
20=-54°51' 
0=-27°25' 
tan 6= -0.519 
K tan #=-9.835 
7i = 42.12 + 9.835 =51.96 
h = 15.44 -9.835 = 5.60 

so that h 2 max= — : — =6.02 

8.625 

and A: 2 2 min= ^?°- =0.649 

8.625 

Hence &i = 2.45, and & 2 = 0.81. 

164. The relation of M to I. In computing the strength 
(moment of resistance) of a beam, it must not be forgotton that it 
is not enough to know the moments of inertia. Suppose we have, 
as is the case of the Z-bar, the max. I x and the min. I 2 . The ratio of 
strength is 

Mi ci cd\ , x 

— = = ^- 1 - (22 

— 1 2 

c 2 

This shows that the strength varies directly as the I, and inversely as 
the extreme fiber distance. For the Z-bar 

Mj _ IiJ P'N' 51.96 P'N' 
M 2 Y 2 PN ~ 5.60* PN ' 

Now, from Fig. 166, PN= - cos 6+ (a + -) sin # = 4.10 

2 V 2/ 

P'N'=-&m0+(a+-) cos 0=1.71 

2 V 2/ 



136 Z-BAR COLUMNS 

HenCe Mt 51.96 1.71 88.85 „ 

— = = - — = 3.9 nearly. 

M 2 5.60 4.10 22.96 

165. Z-bars are used as purlins in the framing of roofs, the 
flanges taking the slope of the roof. Z-bars are used in the construc- 
tion of columns and struts, especially where it is 
necessary to have ready access to all surfaces for the 



-n- 



purpose of painting. The method of using them is 

shown by Fig. 167, which gives a 

general horizontal section. f- — ££- 

The combination of 4 Z-bars, . — 
united by an interior plate, makes 



Fig. 167 

an open column whose principal 




moments of resistance are readily calculated, as 
the axes are parallel to those already used. The 
central connecting plate has usually the same " Fig .. 168 
thickness as the Z-bars. 

In large columns, two wide thick plates are riveted to the outer 
flanges of the Z-bars, as shown in the drawing, Fig. 168. There is, 
however, a slight lack of economy in the use of material as compared 
with some other forms. 



CHAPTER XI. 

Elementary Graphical Statics. 

166. Direct stresses in the members of a frame. 

1. The frames considered in this chapter are ideal; they are as- 
sumed to be rigid; all members lie in a vertical plane; all joints are by 
means of pins which are central to the connected bars; and the only 
stresses (unless otherwise stated) taken into account are the direct 
stresses due to concentrated imposed loads. All loads are applied at 
pins, and their vectors are external to the frames; that is, a load on 
a lower pin is represented by an arrow below the pin; if on an upper 
pin the arrow is above the pin. Every pin is in equilibrium under the 
action of a system of balanced forces, and in every case the stress 
diagram will show for the balanced forces a static polygon. 

2. A bar in tension will be called a "tie"; one in compression will 
be called a "strut." It should go without saying that a tie pulls 
on both pins, and a strut pushes on both pins. All drawings of frames 
and diagrams should be executed on a drawing board, with good in- 
struments and on a scale at least twice that used in this book. Great 



THE STATIC POLYGON 



137 




pains must be taken in drawing parallel lines. In laying off load 
vectors and in measuring stress vectors, the same scale must be used, 
but the scale itself is entirely arbitrary, tons, lbs., kilos, poundals 
or dynes. 

167. Review static polygons. At this point the student will do 
well to review (TO) and make sure that he can draw the static 
polygon for a set of co-planar forces, whose lines of action meet at 
a point and balance, provided all directions are given and all magni- 
tudes but two are known. 

It is also necessary to recall the fact that the arrows on the lines of 
a static polygon show the 
directions in which the 
members meeting at a 
pin act upon it, so as to 
balance. For example, 
suppose the bars 1, 2, 3, 
4, 5, Fig. 169, act upon 
a pin at P, and their 
actions are accurately 
shown by the force poly- 
gon which closes, as the 
arrows follow each other 

around the Area P. These arrows show that Nos. 1, 3 and 5 must 
be struts; and that Nos. 2 and 4 are ties. The student will do well 
to study the figure 169 carefully so as to see the system followed. 

168. Given an irregular cantilever frame supporting at different 
pins six equal loads. Fig. 170. 

1. We begin with the pin A, because that is the only one whose 
static polygon (triangle) can at first be drawn. The forces are the 
weight Wi, and the actions of a and b; Fig. (b) shows the static tri- 
angle, QPN, whose area is also lettered A. Taking the forces in clock- 
wise order round the pin, i. e., Wi, a, b, the triangle shows that a 
acts up against the pin, and is therefore a strut; and that b acts to the 
left, from the pin, being therefore a tie. 

2. The next pin is B acted upon by four forces: a already found 
to be a strut acting down on B ; a known load, W 2 , and the bars c and d. 
Taking the known forces in clockwise order, and using the stress line, 
NP, for a as drawn, we add W 2 = PS; then a line, ST, parallel to C, 
and get back to our starting point, N, at the upper end of a, by a line 
TN parallel to d. We thus have for the pin B the static polygon: 
a _ jy 2 _ c — d, or NPS T — N. The polygon should be clearly identified 



Fig. 169 




138 



GRAPHICAL STATICS 



no matter what other lines are in the neighborhood. It is seen that 
c is a strut, and d is a tie, 

3. We next consider the pin C, acted upon by two known, and 
two unknown forces. Starting at Q, we trace b — d, and close back 
to Q by e -f. The reader must clearly identify the polygon b — d — e—f, 
or QNTR — Q, thinking of the directions of the actions as he goes 

around: e is 
down, and / is 
up. All the bars 
acting on the 
pin C are ties. 

4. Five forces 
act on D, of 
which three are 
known. Begin- 
ning with e at 
the point R, we 
retrace e — c, lay 
down W z , draw 
g and In, and we 
have the poly- 
gon/). Remem- 
ber that every 
static polygon 
must close. 

5. In draw- 
ing polygon for 
E, we must 
begin with W*, 
since the known 
forces must be 
taken first in 
clockwise order; 
and as / must 
follow Wa, the 
latter must be 

laid off above W±\ hence we have W^-f-h-i-j, the last two being 
determined by required directions, and the necessity of returning to 
the point S. 

6. The polygon for F is readily drawn, beginning and ending with 
the point T. 

7. The four-sided polygon for G: j-l-m-n, being a quadri- 
lateral with an internal intersection, is readily seen in Fig. (6). 




A STRESS DIAGRAM 139 

8. The reader should re-draw the figure from the frame alone, 
identify every polygon, and note the nature of the stress in every bar. 
If he puts an additional load at G, he will find that the stress lines for 
p and j will not overlap. 

169. A careful study of the above exposition will bring to the 
reader's notice several important relations between Fig. 170 and 
Fig. 1706. 

1. We followed a clockivise order in considering the forces acting 
on a pin; as for example, the pin /; we had four known forces; so we 
took them thus: o — m — h — W& — q — r: right-handed; in identifying 
the polygon, we took the stresses in the same order, but we went about 
the area I in a left-handed direction. 

2. Not only does every pin or point of intersection in the frame 
have an area in Fig. (6), but every terminal meeting point in (b) has an 
area in Fig. 170. For example, take the point X, in (6) where the stress 
lines of Z, m, h meet; in Fig. 170 the bars Z, m, Jc surround the area X. 

3. Again, we see that at the point Q, in Fig. (b), the force lines of 
W 4 , b,f, and W\ meet; now, turning to Fig. 170 we see that the bars f, b, 
and the indefinitely extended load lines, W\ and W^ bound, or shut off, 
the indefinite area represented by Q in Fig. 170. Similarly, the area 
Y, Fig. 170, corresponds to the meeting of the boundary lines W*>, k 
and W§. 

4. We note that the bar k, in Fig. 170, is the line separating the areas 
X and Y (like a division fence); it is evident that the bar k might be 
read: "the bar XY" (like the Mo. -Kansas line); and the stress in that 
bar as ZF in Fig. 6. 

5. It will be seen that the external forces, i. e., the loads W±, Wi, 

Wi, We, all fall in the same vertical line in (b), and in 

regular order as one passes around the frame externally in a clock- 
wise manner. The line of loads is called, "The Load Line." 

6. If all the areas, within and without the given loaded frame, 
were suitably lettered, any pin could be identified by naming the 
areas around it; thus the pin A could be read: "the pin QPN," which 
is the way we read the static triangle of the forces at that pin. 

The last two suggestions will be adopted hereafter in lettering 
frames and stress diagrams. 

1TO. Method of lettering. A single example will show its 
advantages. 

Fig. 171 shows a second example of an irregular cantilever frame 
with a definite load at each pin. All areas are lettered both within 
and without, and the order of procedure at every pin is indicated 
by a curved arrow. The stress diagram is shown in Fig. b. The 



140 



GRAPHICAL STATICS 



reader will readily see in Fig. 171 the bars CD and AH; and in (6) the 
magnitude of the stress in CD and again in AH. If he wishes to 

know the nature of the stress 
in CD, he must think of its 
action on one of the pins. If 
he thinks of the upper pin, he 
will read that pin (that is the 
areas around it) as G—C—D— 
E — F; now he will read the 
stress polygon also as G — C — 
D — E — F, and note that he 
read C — D downwards; hence 
the action on the upper pin is 
down; so CD is a tie. Had 
he thought of, and read, the 
lower pin, he would have 
(mentally) said D — C, which 
reads up on the stress; so it 
pulls at the lower end as well. 
In the same way he finds that 
ED is a strut. 

171. Cantilever truss. Draw a stress diagram and scale off the 
stresses for the bars shown in Fig. 172. 

172. Roof Trusses. 

Ex. 1. Plain Roof Truss, with 
equal loads on all upper pins. 
Fig. 173. By symmetry the supports 
are equal. Directions: Draw first the 
Load Line, lay off the loads and 
supports to scale and letter them 
from K round to X. Next draw the 
static triangles, XAR and XKJ. 
Next, the static polygons QRAB 





Fig. 172 




and BAXC on the right, 
and JKLI and XJIH on 
the left. Finally, draw for 
the remaining upper pins. 
The symmetry of the figure 
will be an aid to accuracy. 
The many sided polygon 
5 for the central lower pin will 
be found already drawn. 



UNSTABLE FRAMES 



141 



The drawing will be finished when the character of every member is 
shown by the letter 5 (strut) or t (tie). 

Ex. 2. A Hip Roof. Fig. 174. The points A and D in (b) coin- 
cide. Point out the polygon 
for the pin near E in the figure, 
174 (6). 

Ex. 3. Invent a roof truss 
having only triangular areas in 
the frame, and place on it sym- 
metrical loads. Letter and draw 
the stress-diagram, and scale it. 

173. 





Before giving more 
said about unstable 



Pig. 174 




Unstable frames. 

problems, a word must be 

frames. 

1. Suppose a frame with pin joints is hinged at 

X and Z and 

loaded as shown 

in Fig. 175. [The 

reader of what 

follows must draw 

the stress-diagram 

as he reads, or he 
will not appreciate the thought.] To 
find the stresses on the assumption 
of stability, we lay off the loads W\> W 2 and Wz\ then begin with the 
pin CD A. We can next go to either the upper pin or the lower as in 
each case there are only two unknowns. Going to the upper, we say: 
CABE — C and draw AB and BE. Now, going to the lower pin, we 
say: BADF — B; and the only unknown is FB; but, if it be drawn 
in the direction given, the polygon will not close; hence the frame is not 
statical, and cannot support a load. 

If a diagonal bar from Z is put in, it will divide the area, and the poly- 
gon will close with a stress line BP, the length of which stress line shows 
how much a diagonal tie was needed. Equally well (and sometimes 
better) the other diagonal could be put in as a strut. It is thus seen 
that a four-sided area in a frame is unstable with pin joints. Even 
when the joints are nailed, screwed, glued, bolted or pinioned, they 
lack the stability which a diagonal strut or tie readily gives. A 
diagonal wire with its ends well secured is better than much nailing 
and glueing. Notice the weakness of screen frames, tables and benches 
when lateral forces act, and the sagging of doors and gates without 
diagonals when heavily loaded. 



142 GRAPHICAL STATICS 

2. A rectangular frame without a diagonal is made rigid only by a 
large expenditure of material and labor on the joints. The economy 
and safety gained by the use of struts or metallic ties are worthy of 
more consideration by carpenters and cabinet-makers. Note the 
stiffness of aeroplanes. A four-sided frame often may profitably 
have two diagonal ties. A tie rod, wire or bar is a very simple affair, 
requiring little material and little workmanship as compared with a 
strut. 

174. Queries to be answered by drawings. 

1. If only a diagonal tie is put in, would the tension in it be greater 
or less than the compression in the strut, had the other diagonal been 
used? 

2. Suppose the diagonals are of different lengths, the area B not 
being rectangular? Answer as above. 

3. Suppose both diagonals are put in, which would take the stress? 
This is rather "indeterminate"; generally the tie or the one with the 
closest fittings. This query is answered later under Deformations, 
if the fittings are perfect and independent, and all bars are elastic, 
with known dimensions, and the strut does not buckle. 

4. It was seen a few lines above that the diagonal inclined down- 
ward towards the support was a strut, while that inclined upward 
towards the support was a tie. When a vertical frame is subject to 
horizontal forces (in its plane) which may act either way, two diagonal 
ties suffice for both contingencies, and are vastly simpler than a single 
bar built to act as either tie or a strut as may be required. 

Fig. 176 shows a rectangular screen frame, exposed to the action of 
horizontal forces tending to rack the joints. 

When the force at B is acting, the tie, BD, 
supports the frame; when the force at A is act- 
ing, A C is in tension. In either case, if the feet 
are blocked, the action upon the parts DE and 
CF causes them to bend, and develop stresses 
which will be discussed later on. 

175. Roof trusses under symmetrical loads. 

1. By means of "purlins" (girders which cross 
from rafter to rafter), the loads on large roofs are brought to the trusses 
at the joints, or at supported points which might be joints. All loaded 
points, therefore, may be supposed to be joints with pins. 

2. Draw the stress diagram for the loaded roof shown in Fig. 177, 
and mark the character of each member. Though the rafters may 
be continuous from eaves to ridge, they may be supposed to have 




ROOF TRUSSES 



143 




pins where the loads are 
applied. The stress in JK 
is found to be greater than 
in EM. 

3. Figures 178 and 179 




are possible designs of a roof 
truss, which the reader may 
load, letter and draw. 

176. Loads both above and 
below. A roof truss is often 
required to carry loads sus- 
pended from the lower pins. 
In that case the closed polygon 
of loads and supports (external 
forces) is a vertical line, some parts of which are used four times. 

For example, take the simple truss, Fig. 180, in which the lower ties 



Fig. 177 





(called the lower chord) are arranged to give more head-room below 
as shown in the sketch. The "load line" (the external forces taken 

in order around the truss, 
beginning at F) should be 
drawn first. The succession 
of forces is shown by lines 
not quite vertical on the right 
of the proper load line. 
Fig. 180&, on next page. 

177. A Fink truss. It 

often happens that a frame, 
or truss, has its long struts trussed (to prevent bending or sagging) 
by what may be called secondary trusses, thereby creating what to 
a novice seems an impossible case; this is well illustrated in what is 




144 



GRAPHICAL STATICS 



called the "Fink Truss." Fig. 181. It is seen that the rafters, after 
having been bisected, are again sub-divided by small secondary trusses, 

tho they are not wholly inde- 
pendent. For the sake of 
simplicity, uniform loads are 
given. There is no difficulty 
with three pins at either end. 
On going to a fourth pin, one 
finds both roads blocked, 
each pin having three 
unknowns. The difficulty 
may be overcome in different 
ways, but the simplest will 
be for the student to calcu- 
late the stress in XG by the 
method of moments. The 
whole trussed rafter to the 
right support is loaded with 
30 T (centered at the middle 
pin), and is held in equilib- 10 

rium by the support, by XG, 
and by an action at the upper 
end, Z. Taking the upper 
end, Z, as an axis of moments, 
we have the moment equa- 
tion, calling the span s, 




Fig. 180 (b) 



4 2 



0. 



If the height h = - , 

2 V3 

we find that 

XG = 20 V3 = 34.64, 

which is laid off to scale 
from X, Fig. 181(6) deter- 
mining the point G on the 
diagram. Now take the pin 
DCXG. The rest of the 
problem is easy. 

178. Unsymmetrical 
loads. 1. Vertical loads 




THE FINK TRUSS 145 

directly over supports do not affect the stresses in the members of a 
truss, hence they do not appear in the diagram. Pin loads are rarely 
equal, and it is found that when the interior members (the "brac- 
ing") are most severely stressed, the loads are not symmetrical. In 
such cases the supports are unequal and should be calculated (most 
readily by the method of moments), as already illustrated in Chapters 
III and V. See Fig. 182. 

To find the support at one end, take a moment axis at the other end. 
In getting moments, use any 
convenient unit of length; 
the absolu te length of a truss 
is of no importance.* 

2. Ex. Let Z be the axis 




Pig. 182 



for finding F 2 . The upper 
joints divide the span into 
eight equal parts; the lower 
joints into three equal parts. 
Hence, let the span be 24 units; then we have for finding V 2 

24F 2 = 12 + 36 + 36 + 120 + 180 + 252 + 252 + 256 + 64 
F 2 = 50s, and consequently Fi=35§, 

The stress diagram will be found to be unsymmetrical. 

3. There are other ways of finding V\ and F 2 , chiefly by drawing, 
but none are more simple or more readily used; even the above solu- 
tion may be made more simple (and the student should simplify every 
mathematical and mechanical process) by noting that 12 = 4 + 8; 
14 = 6 + 8, and 16 = 8 + 8, so that the loading is equivalent to a sym- 
metrical load which gives 27 J 1 on each wall, a load of 24 J 1 on the center 
pin of the right-hand rafter, and a load of 81* on the right-hand lower 

3 2 

pin. The right-hand support, F 2 , carries -of the 24 T, and - of the ST. 

So that the whole support on the right is 

F 2 = 27 + 18+5i=50i 

all of which is seen without putting pencil to paper. (The student 
should go over this paragraph till every step is clear.) 

179. Inclined loads, wind pressures. When load lines are not 
vertical, the supporting forces, one or both, are not vertical, and the 
external forces acting on a truss form an open static polygon. Some- 



* Given the concentrated loads, the mere size of a truss has nothing to do with 
the direct stresses in its members, tho it has much to do with the way in which 
the members shall be given sufficient strength and stiffness. 



146 GRAPHICAL STATICS 

times the stresses due to vertical loads and those due to inclined loads 
(the wind) are found separately and added. The following illustra- 
tion takes them all together. 

1. Steel trusses are generally anchored firmly to the windward 
wall (on the side which is exposed to the highest winds), while the 
free ends are either resting on rollers, or, placed in grooves or shoes 
which admit of an easy horizontal motion. The end is left free so 
that the truss may expand or contract as the temperature rises or 
falls, or lengthen and shorten again when loads are put on and taken 
off, without moving the walls. Accordingly the supporting force 
at the free end is vertical, or nearly so. 

It is usually assumed that wind pressure is normal to the surface 
of action, and that it is uniformly distributed. Neither assumption 
is very near the truth.* The wind pressure per square foot is a very 
uncertain quantity. It depends chiefly on the velocity of the wind 
and the slope of the roof; and somewhat on the extent of surface. 
Its action is concentrated at the joints as is the "dead" load. 

2. To find the stresses in the members of the roof truss shown in Fig. 
183, ivhich carries dead loads, and wind pressures. The support, V2, is 
found by getting the moment of the wind pressure about an axis at 
Z, the fixed end, and adding the support due to the symmetrical 
dead load. [The drawings are on the opposite page.] 

The support due to the dead load is 12 tons. 

The total wind pressure is 36, and its "center" is at the middle of 
the rafter. If r be the length of the rafter, the wind's moment is 18r. 
Let the inclination of rafter be 30° = /3. 

Then the support, F 2 ' due to the wind, at the wall on the left, has 
a moment arm, 2r cos^S; hence we have the equality of moments 

%V%r cos/3=18r 

V 2 ' = -^— n =6 V3 = 10.39 
cosp 

and V 2 = 12+ F 2 ' = 22.39 

Now, the static polygon for external forces can be drawn, beginning 
with XJ, which is 22.39. Since the polygon must close, the last line, 
SX t shows the magnitude and direction of the support V\. The 
builder must see that this support is furnished. The effect of the 
wind is plainly shown in the stresses of rafters and bracing on the 
windward side, and the chords XA and XC. 



* See a paper on Wind Pressure by Prof. F. E. Nipher, Proceedings of St. 
Louis Academy of Science. 



STRESSES CAUSED BY WIND PRESSURES 



147 



180. Supports modified by wind pressures. The student who 
has read Chapter V will recall that the resultant of vertical and in- 
clined loads could have been found graphically by means of the 
Chain Polygon, and that the 
components of the force 
balancing it, i.e., V\ and V2, 
could also have been found by 
the method there explained. 
However it is doubtful if there 
is a simpler solution than that 
just given, but the student 
should not assume that there 
is no other method. 

Ex. Let the student add 




wind pressures to the load on the truss 
lOwninFig. 177, assuming that thewind- 
ard support is anchored, while the 




other rests on 
rollers. 

The magnitude of possible 
stresses in the bracing should 
determine their design and 
specified strength. It is false 
economy to save material by building a truss 
so light and weak that it fails in a storm. 



148 



GRAPHICAL STATICS 




w w 

Fig. 184 



Bridge Trusses. 

181. The two trusses of a bridge are connected by cross-girders 
which usually rest upon the pin joints of the trusses. These girders 
carry stringers, and the stringers carry the roadway; hence the loads 

are applied to the trusses 
at the joints. The weight 
of a me m b e r itself is 
divided equally between 
its ends. Bridges may 
have "thru" or "trellis" 
spans for moving loads to 

pass between the trusses, or "deck" spans when the roadway and its 

immediate supports are above the trusses. The stresses in members 

are readily found graphically, the 

results may be checked by other 

methods. 

182. A "Warren Girder," or tri- 
angular truss. Fig. 184. The tri- 
angles are equilateral, and only loads FED 
on the lower pins are considered. The 
load line is, in order, 

+JK-KL-LM-MN-NP+PJ. 

The fact that the letters /), E and 
F fall together shows that, under a 
symmertical load, the bars DE and EF 
explained: the stress in FE (tension), 




Fig. 184 (a) 




Fig. 185 



have no stress. This is thus 
caused by the load MN, is 
exactly cancelled by 
the compression caused 
by the equal load LM. 
Were the loads NM 
and ML unequal, both 
FE and ED would be 
under stress. 

2. A deck span. The supports are each 2 W, and the load line 
begins -\-PJ. Fig. 185. With this ideal loading the bars DE and 
EF are idle, but such would not be the case with irregular loading. 
Had the bridge been one panel longer, there would have been no idle 
member. Make a drawing to confirm this. 

3. Draw a stress diagram for a thru span, and one for a deck span, 
with eight or ten panels under uniform loads. Mark struts and ties, 
and take note of laws of increase and decrease of stress in members. 



THE LATTICE GIRDER 



149 




Fig. 186 



183. Combination trusses. When panel lengths are too great 
for roadway stringers, two Warren girders are sometimes combined, 
as shown below. One rests on masonry, as in Fig. 184, and the other 
hangs on piers of steel or masonry as in Figure 185. Such a com- 
bination is known as a Lattice Girder. 

2. Draw a diagram for the suspended truss with its six loads, and 
a separate diagram for the other with its five loads. Fig. 186. 

Scale off all the stresses for the different members, and add the stresses 
in the chords wher- 
ever they have been 
used in both trusses. 
Finally write the 
stresses found on 
the members them- 
selves, as represented 
in the drawing. 

It will be seen that the end panels of both chords are used but once. 
Let the component girders be drawn on good paper with an actual 
span of 12 inches. 

3. Lattice girders are often met with which have three or four 

component trusses. In 
such cases some of the 
end members are 
inclined at special 
angles. See Fig. 187. 

Fig. 187 ^ 

184. Two typical 

bridge trusses are shown in their simplest form in order that the 
student may see their characteristic features, and by personal expe- 
rience may learn the beauty and simplicity of the graphical analysis 
when one has good drawing instruments at hand. 

1. The Pratt Truss, shown in Fig. 188. 

2. The "Double Pratt" has short vertical struts and long diagonal 
ties. The two trusses 
have the same chords 
and end struts, but 
separate diagonals. 
The load on the 
vertical rod, near 

the end, is divided, usually, half and half. See Fig. 189. 

Draw the diagrams separately, measure stresses and where the 
members are superposed, add the stresses. 





Fig. 188 



150 



GRAPHICAL STATICS 




Fig. 189 
A DOUBLE PRATT 




Fig. 190 
A HOWE TRUSS 



185. A Howe truss. If a load is perfectly symmetrical (and it 
rarely is), the dotted members, which are both ties, will not be needed; 

with a moving load both are 
needed, but not both at once. 
The student will note the long 
diagonal struts in the "Howe," 
and the long diagonal ties in 
the "Pratt." The "Howe- 
went out of use when steel replaced wood in bridge construction. 
Fig. 190. 

For a full discussion of "Framed structures" which the above few 
pages are intended to lead 
up to, the student is referred 
to Building Construction 
and Bridge Engineering. 

Calculations base on 
stress diagrams. In all the 
problems of the present chapter, the magnitude of stresses in frame mem- 
bers was determined graphically, and the degree of accuracy depended 
upon the quality of the instruments and the skill of the draftsman in 
laying off and measuring lines and angles. But it is easily seen that 
the diagrams, even when made "free-hand," lend themselves to trig- 
onometric calculation. An examination of any one of the diagrams 
will suffice to discover how an unknown length can be computed, as 
all angles are known. 

1. For example, let the student calculate the stresses in several 
members of the Fink truss, Fig. 181, and also of the Pratt Truss, 
Fig. 188, and compare results with his measurements of his stress- 
diagrams. He is quite likely to find defects in his instruments and 
fault}^ drawing. 

2. Let the student make a stress diagram of the cantilever truss, 

Fig. 42. 

186. The endless variety of designs. There is no limit to the 
number of frames for roofs and bridges. "King Post" and "Queen 
Post" roof trusses offer no difficulty, but no unusual design should 
be built without a careful drawing which shows both the nature and 
the magnitude of the stress in each piece. The writer once saw a 
roof truss for a church actually built with a tie (an eye-bar), where 
there should have been a strut. The builder's "horse-sense" was 
inferior to a stress diagram. A drawing of thetruss follows. 
Fig. 191. 



SCIENCE VS. HORSE-SENSE 



151 



The reader may assume any reasonable loading, and proceed with 
the drawing, using one of the 
members, FG; and note the 
stress in it. Then make a second 
diagram with the other diag- 
onal, FG. See which would 
alone be a tie, and which alone 
would be a strut. The reason 
why the truss does not fall is, 
that the rafter is continuous to 
the ridge and very strong, and 
its stiffness saves it. 




Pig. 191 



CHAPTER XII. 

Internal Stress. 

Composition and Resolution of Stresses. 

187. In Chapter IX it was shown that at every internal point 
of a loaded beam there was stress normal to a cross-section plane, and 
a shearing stress in the same plane. 

The definition that "force is an action between two bodies," is to 
apply to two contiguous parts of a body. 

Let A, B and C be parts of a horizontal loaded beam. Both A and 

C and the load Aw are acting on B, and 
holding it in equilibrium. There are actions 
at every point of the planes which bound 
B, and those actions vary from point to 
point. 

But that is not all. If a prism like that 
at D be formed by three general planes 
cutting the same rectangular loaded beam, 
there is stress (actions) at every point of every lateral face. The 
adjacent material is pressing or pulling D, and is dragging or tending 
to drag it up or down; and yet, as D is at rest, we know that all such 
actions actually balance. The prism may be very small, infinitely small, 
yet have three sides or faces, and the stresses on those sides are normal, 
oblique and tangential. It is the purpose now to study the stresses on 
every plane that can pass thru a given point in a loaded beam, particu- 
larly planes which are perpendicular to the plane of external forces 
(loads and supports). 




Fig. 192 



152 



INTERNAL STRESS 



188. Simple stress and its components. Fig. 193 represents a 
bar in tension but under no other stress. If the cross-sectional area S is 

uniform, the "Normal Intensity" is m .„ 

Px= T / s 

T 

upon every unit of surface of a right section. 
If an oblique section, AC, is made, the stress 
is oblique, and as the section has the area Si = S 
secOy we have 

T 

p x cos u 




P = 



S sec 9 



If now p be resolved into^ its normal and 
tangential components, we have 

Vn~V cos Q~Vx cos 2 9 
p t = p sin 9 = p x sin 9 cos 6 



If a second plane HK be taken so that its inclination to a right sec- 
tion is <f>= — 9, we shall have 

2 

Vn =Vx cos 2 <j> = p x sin 2 9 

p t r = p x sin <f) cos <f) = p x cos 9 sin 9 

Pn+Pn=Px> and ^ = 2?/ if <f) + 9 = 7r/2. 



Hence 



This is particularly important, and is stated as follows: If two 
oblique planes having their normals in the same axial plane are at 
right angles, the tangential components of the oblique 
stresses are equal. A graphical representation of magni- 
tudes of the stresses on an oblique plane is readily 
shown: The value of p n is found by projecting p x twice 
thru the angle 9. Fig. 194. 




Fig. 194 



189. Compound stress. More often than not, 
material is subjected to, or sustains, more than one 
simple stress. When the simple stresses are known, the 
stress on any given plane is readily found if all the 
normals lie in the same plane. This may be illustrated by finding 
the stress on a third plane when the simple stresses on two planes are given. 

1. Take the case of the shell of a steam boiler or a cylindrical gas 
tank. Let the radius of the cylinder be r, the thickness of the shell 
be c, and the gauge pressure of the steam or gas per square inch be 
q in excess of the external air-pressure. The total direct longitudinal 



STRESSES IN THE SHELL OF A STEAM BOILER 



153 



tension is qrrr 2 due to the pressures upon the ends of the cylinder, which 

must be sustained by the shell at every cross-section, whose area is 

A = %irrc. „ 

TT q-rrr 2 or 

Hence p y = - = — 

Qirrc 2c 

if the direct longitudinal tension is p y . 

2. In addition to this longitudinal tension, there is a "hoop tension" 
which is found by considering the tension in a hoop one inch wide 
around the cylinder. The resultant internal pressure on one half of 
the hoop is the same as the pressure would be on a diametral strip one 
inch wide; hence %rXq = %T, where T is the tension at every cross- 
section of the hoop. See Fig. 195. 

Hence T = rq. 

Now, the cross-section of the hoop is cXl, so that 

p x = T/c = rq/c 

This shows that the intensity of the hoop tension is twice the intensity 
of the longitudinal tension.* 

3. Unless otherwise stated as to the drawings which follow, the 
planes thru OX and Y are thought of 
as perpendicular to the plane of the 
paper, and intersecting in an axis OZ 
(not shown); and all oblique planes of 
action are parallel to OZ. 

4. It is now required to find the stress 
on the plane, or at the section AB, 

which makes 
an angle 6 with 
OY. Fig. 196. 
Let ON be 
normal to the 

section. By the formulas just derived, resolv- 
ing p x into components on AB. 

p z cos 2 0, p/ = p z sin cos 0; 





Fig. 196 



Vn 

Resolving p yi 

p n " = p y sin 2 0, 



Pt ~P y sin cos 6 



* The observing student will recall the fact that there are two rows of rivets 
along a longitudinal seam of a steam boiler, but only one row on a cross-sectional 
seam. 



154 



COMBINATION OF INTERNAL STRESSES 



The magnitudes of p n ', p n ", p t ', p" , etc., are found by projection in 

Fig. 197. — 

OX = p x ; OY = p v ; OC = OC' = p n '; OD = OD' = p n ». 

ON=p n = p n '+p n »; 

NR= Pt '-p t » = p t 

0R = Pr 

From the figure it is seen that p n ' and 
Pn have the same sign, while p t f p" 
have opposite signs. 

Hence we have the final equations 

Pn = Pn +Pn" = PxCOS 2 0+p y SlU 2 1 

Vt = Pt' -Vt" = (Px-Py) sinOcosO ) 

If we lay off p n on ON, and p t at 
right angles to ON on the side of the 
larger of p t ' and p" , then p r = OR is the 
stress on the plane AB per unit of 
surface. It is thus seen that the stress 
p r is oblique, and that the obliquity, 
that is, the angle with the normal, is </>, 
and that 

p x cos 2 0-\-p y sin 2 
Substituting for p z and p y the values qrjc and qr/Zc, we have 




p n = — (1 +cos 2 0);p t = — sm ^ cos 
2c 2c 



tan<^> 



sin cos 
l+cos 2 # 



19©. When the stresses on OX and OY are normal and known, 
it is much simpler to use the static triangle. Suppose we wish to find 
the stress on a general plane which is parallel to the inter-section of 
the planes OX and OF, and which makes an angle with OY. Fig. 198. 

1. ON is normal to AB, hence XON = 0. Take an elementary 
prism, whose faces are parallel to the planes AB, OX and OY, whose 
altitude is c, and whose bases have the edges bh = dx, ak = dy, ab = ds. 
It is evident that dy = ds cos 0, and dx = ds sin 0. 



THE STATIC TRIANGLE OF STRESS 



155 




Fig. 198 



Now the face of the prism, akXc, has the stress p x on every unit of 
surface; hence the stress on that face, which we will call OE, and lay 

off on OX, is — a 

OE = cp x cost/ ds 

Similarly, the stress on the face, 
kb, is found to be 

OD = cp y sin ds 

The resultant of these two 
stresses is 

OR = cds (p x 2 cos 2 0+p y 2 sm 2 0) h 

If the intensity of stress on 
the third face is represented by 
p r , the total stress on that face 
is p r cds, but that stress must 
balance the resultant OR. Hence 
the static triangle, OER, and 

RO = p r cds, 

Vr=(Px 2 cos 2 + Py 2 sin 2 #)* 

This formula could easily have been found by reducing p r = {p n 2 -\-p t 2 )' 
in the former solution, 188. 

2. It will be noted that the height of our prism, and the hypotenuse 
ds have disappeared; only unit stresses remain. 
Hence, we will use only unit stresses, and in the 
place of OE = cp z cos 0ds, we will use OF = p x 
cos 0. Fig. 199. 

If we lay off on OX p x cos = OF, and on OY, 
p y sin = OG, and draw the diagonal, that diagonal, 
if drawn towards 0, and read RO, will represent 
in length and direction the unit stress which 
balances the stresses p x and p y . This statement 
is readily grasped if the student bears in mind 
that the stress of adjacent material on ds must 
balance the stresses on the faces dx and dy, since 
the prism is in equilibrium. 
191. The ellipse of stress. Fig. 199. Let AB be the plane on 
which the stress is to be found. Draw the normal, ON, and lay off 
on it both p x = 0H and p y = 0K, each to scale, and project them upon 
their respective axes, thus getting the co-ordinates of R and the mag- 
nitude of p r to scale. 




Fig. 199 



OCT 3 1912 



156 



THE ELLIPSE OF STRESS 



In the case of the steam boiler under internal pressure, it was found 
that p x /p y = %, but that relation was special; in general they may have 
any ratio. Whenever p x and p y are normal, and are the only stresses 
on their respective planes, which are perpendicular to each other, 
they are called the principal stresses, and their planes of action are 
called the principal planes, and the magnitude of p r on any oblique 
plane lies between the magnitudes of p x and p y . From this construc- 
tion it is evident that the locus of R as the plane AB changes its posi- 
tion, is an ellipse, whose semi-major axis is p x > and whose semi-minor 
axis is p y . If we call OF = x and OG = y, we have x = p x cos0, and 
y = PysinO; eliminating 0, we have 



£1 _i_ ML 

Vx Py 2 



= 1. 



The figure is known as the Ellipse of Stress. 

The ''obliquity" of the stress is shown by <j>, and the components of 

v r , ON and NR, are 

1 ON = p n , and NR = p t 

Their analytic values have already been found by projection; 

ON = p n = p x cos 2 0+pysin 1 0, and NR = p t = (p x — p y ) cosOsinO. 

(p x — p u )sinOcos 



and 



tan<£ = 



p x cos 2 O+Py sin 2 6 



192. Caution. 



Unless great care is exercised, the young student 
will overlook an important matter in thinking 
about finding the component stresses on an oblique 
plane, when p z and p y are given as the stresses 
on two rectangular planes. 

The temptation is to resolve p x into p n ' and 
p t ', and p y into p n " and p" as shown in Fig. 200, 
which is all wrong. 

It is not the unit stress, p Xi on the plane OY, 
which is to be resolved into its components; it is 
the unit stress on the plane AB which is to be 

resolved. That stress is not p x , it is p x cos0> and that is resolved into 

p x cos 2 6 = p n ', and p x cos 6 sin 9 = p t f . 

193. 1. The practical value of a problem like that of a boiler 
shell lies in the fact that it shows at a glance that when for any reason 
an oblique section in the shell is made, and the adjacent plates are 
connected by butt-straps, bolts or stays the direction of the bolts or 




KSffi 0F CONGRESS 




